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\begin{document}

\begin{center}
\textbf{Hopf algebras, from basics to applications to renormalization}

\textit{Dominique Manchon}

Revised and updated version, may 2006

arXiv:math/0408405v2

\textbf{Errata and questions (version 2)}

\bigskip
\end{center}

\begin{itemize}
\item There are two different references called [EGK1].

\item \textbf{Introduction:} In the second line of page 2, you have a
redundant comma ("element of $\mathcal{A}_{-}$, , and $\varphi_{+}$ is a map").

\item \textbf{I.1, before Proposition I.1.1:} You write: "for any bilinear map
$f$ from $A\times B$ into $C$". I would consider replacing the "into" by a
"to" here, since some people read "into" as "injective".

\item \textbf{Proof of Proposition I.1.2:} You claim that the map $j$ "is
easily seen to be injective". While the proof of this injectivity indeed looks
easy when one reads it (see e. g.
http://mathoverflow.net/questions/72013/homa-c-homb-d-injects-into-homab-cd-when-why
), I am not sure whether this proof is that easy to come up with on one's own.
You might want to give a few hints here...

\item \textbf{I.2.1:} The definitions of a left ideal and of a right ideal
should be interchanged: You define a subspace $J\subseteq A$ to be a left
ideal if $m\left(  J\otimes A\right)  $ is included in $J$, and a right ideal
if $m\left(  A\otimes J\right)  $ is included in $J$; but this should be
exactly the other way round. (Besides, there is a closing bracket missing
between "$m\left(  J\otimes A+A\otimes J\right)  $" and "is included in $J$".)

\item \textbf{I.2.3:} You write: "A \textit{left }$A$\textit{-module} is a
$k$-vector space $M$ together with a map". Maybe replace "map" by "$k$-linear
map" here, unless you think this is clear to the reader anyway.

\item \textbf{I.2.3:} In the second commutative diagram on page 7, replace the
$A$ in the lower right corner by an $M$.

\item \textbf{I.2.3:} You write: "A left module $M$ is \textit{simple} if it
does not contain any submodule different from $\left\{  0\right\}  $ or $M$
itself." You should add "and is nonzero" here, because otherwise the trivial
module $\left\{  0\right\}  $ would be a simple module, making many of the
following results wrong.

\item \textbf{Proof of Proposition I.2.1:} You write: "and by simplicity of
$M$ the map :%
\begin{align*}
\phi_{m}:A  &  \rightarrow M\\
a  &  \mapsto a.m
\end{align*}
gives rise to an morphism of left $A$-modules from $A/J_{m}$ onto $M$". I
don't think you use the simplicity of $M$ here - unless the "onto $M$" part
means that this map is surjective (but it is useless to state this here, since
you state it again one line below). Thus I would propose removing "by
simplicity of $M$" from this sentence, and replacing the "onto" by a "to".
Also, "an morphism" should be "a morphism".

\item \textbf{Proof of Proposition I.2.2:} Here you write: "write (thanks to
semi-simplicity) $M=N\oplus T$ where $T$ is another $A$-submodule of $M$".
This is indeed a consequence of semi-simplicity, but not of the way you have
defined semi-simplicity! You defined a module to be \textit{semi-simple} if it
can be written as a direct sum of simple modules. But what you use here is
that a module is semi-simple iff every submodule of it is a direct addend. The
equivalence is not completely trivial (although not too hard).

\item \textbf{Corollary I.2.3:} Some people might misinterpret the word "into"
as a statement that the map is injective (which is wrong). I would replace
this word by "to".

\item \textbf{I.2.4:} You write: "The \textit{radical} $\operatorname*{rad}M$
of a left module is by definition [...]". I would replace "a left module" by
"a left module $M$" here (otherwise, the letter $M$ is never defined).

\item \textbf{Remark at the end of I.2.4:} Here you write "Jacobson ideal"
twice. This should be "Jacobson radical". Also, replace $\operatorname*{Rad}A$
by $\operatorname*{rad}A$ (to make the notation compatible with the rest of
your text).

\item \textbf{Proof of Lemma I.2.9:} Replace "for any $i\in\left\{
0,...,n\right\}  $" by "for any $i\in\left\{  1,...,n\right\}  $".

\item \textbf{Proof of Proposition I.2.8:} Replace "a finite-dimensional
primitive ideal" by "a finite-\underline{co}dimensional primitive ideal".

\item \textbf{Proof of Proposition I.2.8:} Here you write: "But $A_{M}%
^{\prime\prime}$ is a matrix algebra over $D$". I don't find this that obvious
- what you are using here is that $M$ is a finite-dimensional $D$-module, and
every finite-dimensional $D$-module is free (which is because $D$ is a skew
field, and because Gaussian elimination and most of the linear algebra based
on it work over skew fields just as well as over fields), so that $M$ is free.

\item \textbf{Proof of Proposition I.2.8:} Replace "according to lemma I.2.7"
by "according to lemma I.2.9".

\item \textbf{I.3.1:} On the first line of I.3.1, you write: "Coalgebras are
objects wich [...]". There is an obvious typo here.

\item \textbf{I.3.1:} Between the two commutative diagrams on page 11, you
write: "Coalgebra $C$ is \textit{co-unital} if moreover there is a co-unit
$\varepsilon$ such that the following diagram commutes :". For the sake of
completeness, I would replace "co-unit $\varepsilon$" by "co-unit
$\varepsilon:C\rightarrow k$" here.

\item \textbf{I.3.1:} In the definition of subcoalgebras (as well as left
coideals, right coideals and two-sided coideals), you write: "is contained in
$J\otimes J$ (resp. $C\otimes J$, $J\otimes C$, $J\otimes C+C\otimes J$) is
included in $J$". Clearly, the "is included in $J$" part of this sentence
should be removed.

\item \textbf{Proposition I.3.1, 1):} I don't think the product of $C^{\ast}$
is really the "transpose" of the coproduct of $C$. The coproduct of $C$ is
$\Delta:C\rightarrow C\otimes C$, and thus its transpose is $\Delta^{\ast
}:\left(  C\otimes C\right)  ^{\ast}\rightarrow C^{\ast}$. To get the product
of $C^{\ast}$, we have to compose this with the injection $C^{\ast}\otimes
C^{\ast}\rightarrow\left(  C\otimes C\right)  ^{\ast}$.

\item \textbf{Proposition I.3.1, 1):} Replace "co-unity" by "co-unit".

\item \textbf{Proof of Proposition I.3.1:} Replace "$\Delta x\subset J\otimes
C+C\otimes J$" by "$\Delta x\in J\otimes C+C\otimes J$".

\item \textbf{Proposition I.3.2:} I am not sure about this, but I think that
this proposition is false. More precisely, all the "if" parts are correct (cf.
Sweedler, Proposition 1.4.3 b) and further), but the "only if" parts are not
(or at least they don't seem correct to me). Also I think the words "Dually we
have the following" before this proposition are misleading - this proposition
does not follow from Proposition I.3.1 by duality.

I think I have a counterexample to the "only if" part: Let $C$ be a connected
filtered coalgebra with $\operatorname*{Prim}C$ (the space of primitive
elements of $C$) infinite-dimensional (for instance, take $C$ to be the tensor
Hopf algebra of a vector space of dimension $\geq2$, or the shuffle Hopf
algebra of an infinite-dimensional vector space). Let $\left(  e_{i}\right)
_{i\in J}$ be a basis of $\operatorname*{Prim}C$, let $x$ be an object not in
$J$, and let $e_{x}$ be the unity $1$ of $C$ (of course, $C$, being connected
filtered, has a unity). Then, $\left(  e_{i}\right)  _{i\in J\cup\left\{
x\right\}  }$ is a basis of the subspace $\operatorname*{Prim}C+k\cdot1$ of
$C$. Extend this basis to a basis $\left(  e_{i}\right)  _{i\in I}$ of $C$
(with $I\supseteq J\cup\left\{  x\right\}  $). Now define a $g_{i}\in C$ for
every $i\in I$ as follows:%
\[
g_{i}=\left\{
\begin{array}
[c]{l}%
1,\text{ if }i=x\text{ (in this case, }e_{i}=1\text{ as well);}\\
1+e_{i},\text{ if }i\neq x
\end{array}
\right.  .
\]
It is easy to see that $\left(  g_{i}\right)  _{i\in I}$ is still a basis of
$C$. Now, let $\left(  f_{i}\right)  _{i\in I}$ be the dual "basis" of
$C^{\ast}$ to the basis $\left(  g_{i}\right)  _{i\in I}$ of $C$ (this means
that $f_{i}$ is the projection on the $g_{i}$-coordinate for every $i\in I$);
of course,$\ \left(  f_{i}\right)  _{i\in I}$ is not really a basis, but at
least a linearly independent subset.

Now define a subspace $K$ of $C^{\ast}$ by $K=\left\langle f_{i}\mid i\in
I\right\rangle $. Then, clearly, $K^{\perp}=0$ is a two-sided coideal of $C$
(which also satisfies $\varepsilon\left(  K^{\perp}\right)  =0$, but this
doesn't even matter, since you don't require coideals to satisfy
$\varepsilon\left(  K^{\perp}\right)  =0$). However, $K$ is not a subalgebra.
This is seen as follows:

Every $i\in J$ satisfies $\Delta\left(  g_{i}\right)  =g_{x}\otimes
g_{i}+g_{i}\otimes g_{x}-g_{x}\otimes g_{x}$ (in fact, this is just another
way to state $\Delta\left(  1+e_{i}\right)  =1\otimes\left(  1+e_{i}\right)
+\left(  1+e_{i}\right)  \otimes1-1\otimes1$, which in turn is just another
way to say that $e_{i}$ is primitive). Thus, $\left(  f_{x}\ast f_{x}\right)
\left(  g_{i}\right)  =-1$. Since this holds for every $i\in J$, and $J$ is
infinite (because $\operatorname*{Prim}C$ is infinite-dimensional), this shows
that $f_{x}\ast f_{x}$ cannot lie in $K$ (since $K$ is the space of all linear
maps $C\rightarrow k$ which are finite linear combinations of coordinate
maps). This means that $K$ is not a subalgebra.

Or is it? I don't feel particularly sure of any counterexamples I produce, as
I know that 50\% of them are wrong.

\item \textbf{Proposition I.3.2:} Replace "rightt" by "right".

\item \textbf{I.3.1:} I believe that the paragraph directly after Proposition
I.3.2 (this is the paragraph beginning with "The linear dual $\left(  C\otimes
C\right)  ^{\ast}$ naturally contains [...]" and ending with "[...] implies
that $u$ is a unit") should rather be placed before Proposition I.3.1. It
defines the algebra $C^{\ast}$ used in Propositions I.3.1 and I.3.2.

\item \textbf{I.3.1:} In the definition of the tensor product of two
coalgebras (in the very last paragraph of page 13), you write: "Let $C$ and
$D$ be unital $k$-coalgebras". The "unital" should be "co-unital" here. Also,
in the same paragraph, "co-unity" should be "co-unit".

\item \textbf{I.3.2:} When defining the notion of a subcomodule, you write:
"$\Phi\left(  C\right)  \subset C\otimes N$". This should be $\Phi\left(
N\right)  \subset C\otimes N$.

\item \textbf{I.3.2:} In the middle of page 14, the formula%
\[
\left(  \Phi\otimes I\right)  \circ\Phi\left(  m\right)  =\sum_{\left(
x\right)  }m_{1:1}\otimes m_{1:2}\otimes m_{0}=\sum_{\left(  m\right)  }%
m_{1}\otimes m_{0:1}\otimes m_{0:0}=\left(  I\otimes\Delta\right)  \circ
\Phi\left(  m\right)
\]
has three typos. It should be%
\[
\left(  \Delta\otimes I\right)  \circ\Phi\left(  m\right)  =\sum_{\left(
m\right)  }m_{1:1}\otimes m_{1:2}\otimes m_{0}=\sum_{\left(  m\right)  }%
m_{1}\otimes m_{0:1}\otimes m_{0:0}=\left(  I\otimes\Phi\right)  \circ
\Phi\left(  m\right)  .
\]


\item \textbf{Proposition I.3.3:} "if and only is" should be "if and only if".

\item \textbf{Proof of Theorem I.3.4:} In this proof, you seem to assume in
that $M$ is a right comodule (rather than a left one). (There is only one
exception: that is when you write "Let us show that $N$ is a left subcomodule
of $M$".)

\item \textbf{Proof of Theorem I.3.6:} Here you write: "$E=C^{\ast}/N^{\perp}$
is a finite-dimensional left module over $C^{\perp}$". Clearly you mean
$C^{\ast}$ instead of $C^{\perp}$.

\item \textbf{Proof of Lemma I.3.9:} In the formula which defines the form
$f_{\gamma}$, replace the word "si" by "if" (two times).

\item \textbf{Proof of Lemma I.3.9:} At the very end of this proof, replace
"$y_{\gamma}$ in in $D$" by "$y_{\gamma}$ is in $D$".

\item \textbf{Proof of Proposition I.3.10:} In this proof you seem to use that
$R$ is the \textit{direct} sum of the simple subcoalgebras of $C$. Why is that
obvious? In my opinion, this requires a further lemma: that any sum of
pairwise distinct simple subcoalgebras of $C$ must be a direct sum. This, in
turn, is a particular case of another theorem\footnote{which can be seen as a
dual to the Chinese Remainder Theorem}: that any sum of subcoalgebras of $C$
all of whose pairwise intersections are $0$ must be a direct sum. This theorem
is proven by reducing it to the case of finitely many subcoalgebras, and then
proving it by induction over the number of subcoalgebras (using Lemma I.3.9 in
the induction step).

\item \textbf{Proof of Proposition I.3.11:} Replace "(lemma I.2.8)" by
"(corollary I.2.10)".

\item \textbf{Proof of Proposition I.3.12:} I don't think the equality%
\[
\left(  X\wedge Y\right)  \wedge Z=X\wedge\left(  Y\wedge Z\right)  =\left(
X^{\perp}Y^{\perp}Z^{\perp}\right)  ^{\perp}%
\]
is that much obvious. The definition only yields $\left(  X\wedge Y\right)
\wedge Z=\left(  \left(  \left(  X^{\perp}Y^{\perp}\right)  ^{\perp}\right)
^{\perp}Z^{\perp}\right)  ^{\perp}$ and $X\wedge\left(  Y\wedge Z\right)
=\left(  X^{\perp}\left(  \left(  Y^{\perp}Z^{\perp}\right)  ^{\perp}\right)
^{\perp}\right)  ^{\perp}$, but it is not clear why these things are the same
as $\left(  X^{\perp}Y^{\perp}Z^{\perp}\right)  ^{\perp}$. Am I missing
something?\newline Here is how I would prove Proposition I.3.12 1):\newline
Let $\pi_{X}$, $\pi_{Y}$ and $\pi_{Z}$ be the canonical projections from $C$
onto $C\diagup X$, $C\diagup Y$ and $C\diagup Z$, respectively. Let
$\pi_{X\wedge Y}$ be the canonical projection from $C$ onto $C\diagup\left(
X\wedge Y\right)  $.\newline There is a well-known fact in linear algebra that
if $A$, $B$, $A^{\prime}$ and $B^{\prime}$ are four vector spaces and
$f:A\rightarrow A^{\prime}$ and $g:B\rightarrow B^{\prime}$ are two linear
maps, then $\operatorname*{Ker}\left(  f\otimes g\right)  =\left(
\operatorname*{Ker}f\right)  \otimes B+A\otimes\left(  \operatorname*{Ker}%
g\right)  $. Applied to $C$, $C$, $C\diagup X$, $C\diagup Y$, $\pi_{X}$ and
$\pi_{Y}$ in lieu of $A$, $B$, $A^{\prime}$, $B^{\prime}$, $f$ and $g$, this
yields $\operatorname*{Ker}\left(  \pi_{X}\otimes\pi_{Y}\right)
=\underbrace{\left(  \operatorname*{Ker}\pi_{X}\right)  }_{=X}\otimes
C+C\otimes\underbrace{\left(  \operatorname*{Ker}\pi_{Y}\right)  }%
_{=Y}=X\otimes C+C\otimes Y$.\newline The definition of $X\wedge Y$ rewrites
as $X\wedge Y=\Delta^{-1}\left(  X\otimes C+C\otimes Y\right)  $. Thus,%
\[
X\wedge Y=\Delta^{-1}\left(  \underbrace{X\otimes C+C\otimes Y}%
_{=\operatorname*{Ker}\left(  \pi_{X}\otimes\pi_{Y}\right)  }\right)
=\Delta^{-1}\left(  \operatorname*{Ker}\left(  \pi_{X}\otimes\pi_{Y}\right)
\right)  =\operatorname*{Ker}\left(  \left(  \pi_{X}\otimes\pi_{Y}\right)
\circ\Delta\right)  .
\]
Thus the map $\left(  \pi_{X}\otimes\pi_{Y}\right)  \circ\Delta:C\rightarrow
\left(  C\diagup X\right)  \otimes\left(  C\diagup Y\right)  $ factors through
$C\diagup\left(  X\wedge Y\right)  $. In other words, there exists a map
$\overline{\Delta}:C\diagup\left(  X\wedge Y\right)  \rightarrow\left(
C\diagup X\right)  \otimes\left(  C\diagup Y\right)  $ such that $\left(
\pi_{X}\otimes\pi_{Y}\right)  \circ\Delta=\overline{\Delta}\circ\pi_{X\wedge
Y}$. Moreover, this map $\overline{\Delta}$ is
injective\footnote{\textit{Proof.} Let $\varphi\in C\diagup\left(  X\wedge
Y\right)  $ be such that $\overline{\Delta}\varphi=0$. Then, $\varphi
=\pi_{X\wedge Y}\left(  c\right)  $ for some $c\in C$ (since $\pi_{X\wedge Y}$
is surjective). Hence,%
\[
\overline{\Delta}\varphi=\overline{\Delta}\left(  \pi_{X\wedge Y}\left(
c\right)  \right)  =\underbrace{\left(  \overline{\Delta}\circ\pi_{X\wedge
Y}\right)  }_{=\left(  \pi_{X}\otimes\pi_{Y}\right)  \circ\Delta}\left(
c\right)  =\left(  \left(  \pi_{X}\otimes\pi_{Y}\right)  \circ\Delta\right)
\left(  c\right)  .
\]
Thus, $\overline{\Delta}\varphi=0$ becomes $\left(  \left(  \pi_{X}\otimes
\pi_{Y}\right)  \circ\Delta\right)  \left(  c\right)  =0$ and hence
$c\in\operatorname*{Ker}\left(  \left(  \pi_{X}\otimes\pi_{Y}\right)
\circ\Delta\right)  =X\wedge Y=\operatorname*{Ker}\pi_{X\wedge Y}$, so that
$\pi_{X\wedge Y}\left(  c\right)  =0$. Thus, $\varphi=\pi_{X\wedge Y}\left(
c\right)  =0$. We have therefore shown that every $\varphi\in C\diagup\left(
X\wedge Y\right)  $ such that $\overline{\Delta}\varphi=0$ satisfies
$\varphi=0$. Thus, $\overline{\Delta}$ is injective.}. The map $\overline
{\Delta}\otimes\operatorname*{id}:\left(  C\diagup\left(  X\wedge Y\right)
\right)  \otimes\left(  C\diagup Z\right)  \rightarrow\left(  C\diagup
X\right)  \otimes\left(  C\diagup Y\right)  \otimes\left(  C\diagup Z\right)
$ is therefore also injective.\newline The diagram%
\[%
%TCIMACRO{\TeXButton{xymatrix}{\xymatrixcolsep{5pc}
%\xymatrix{
%C \ar[rd]^{\Delta} & & \\
%& C \otimes C \ar[r]^-{\pi_{X\wedge Y} \otimes\pi_Z} \ar[d]^-{\Delta\otimes
%\id}
%& \left(C \diagup\left(X\wedge Y\right)\right) \otimes\left(C\diagup
%Z\right) \ar[d]^{\overline{\Delta} \otimes\id} \\
%& C \otimes C \otimes C \ar[r]^-{\pi_X \otimes\pi_Y \otimes\pi_Z}
%& \left(C\diagup X\right) \otimes\left(C\diagup Y\right) \otimes\left(C\diagup
%Z\right)
%}}}%
%BeginExpansion
\xymatrixcolsep{5pc}
\xymatrix{
C \ar[rd]^{\Delta} & & \\
& C \otimes C \ar[r]^-{\pi_{X\wedge Y} \otimes\pi_Z} \ar[d]^-{\Delta\otimes
\id}
& \left(C \diagup\left(X\wedge Y\right)\right) \otimes\left(C\diagup
Z\right) \ar[d]^{\overline{\Delta} \otimes\id} \\
& C \otimes C \otimes C \ar[r]^-{\pi_X \otimes\pi_Y \otimes\pi_Z}
& \left(C\diagup X\right) \otimes\left(C\diagup Y\right) \otimes\left(C\diagup
Z\right)
}%
%EndExpansion
\]
commutes. Hence,%
\begin{align*}
&  \operatorname*{Ker}\left(  \left(  \pi_{X}\otimes\pi_{Y}\otimes\pi
_{Z}\right)  \circ\left(  \Delta\otimes\operatorname*{id}\right)  \circ
\Delta\right) \\
&  =\operatorname*{Ker}\left(  \left(  \overline{\Delta}\otimes
\operatorname*{id}\right)  \circ\left(  \pi_{X\wedge Y}\otimes\pi_{Z}\right)
\circ\Delta\right)  =\operatorname*{Ker}\left(  \left(  \pi_{X\wedge Y}%
\otimes\pi_{Z}\right)  \circ\Delta\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{since the map }\overline{\Delta}\otimes\operatorname*{id}\text{ is
injective, and thus composing with it}\\
\text{does not change the kernel}%
\end{array}
\right)  .
\end{align*}
Now, applying $X\wedge Y=\operatorname*{Ker}\left(  \left(  \pi_{X}\otimes
\pi_{Y}\right)  \circ\Delta\right)  $ to $X\wedge Y$ and $Z$ instead of $X$
and $Y$, we get $\left(  X\wedge Y\right)  \wedge Z=\operatorname*{Ker}\left(
\left(  \pi_{X\wedge Y}\otimes\pi_{Z}\right)  \circ\Delta\right)  $. We
conclude that%
\[
\operatorname*{Ker}\left(  \left(  \pi_{X}\otimes\pi_{Y}\otimes\pi_{Z}\right)
\circ\left(  \Delta\otimes\operatorname*{id}\right)  \circ\Delta\right)
=\operatorname*{Ker}\left(  \left(  \pi_{X\wedge Y}\otimes\pi_{Z}\right)
\circ\Delta\right)  =\left(  X\wedge Y\right)  \wedge Z.
\]
Similarly,%
\[
\operatorname*{Ker}\left(  \left(  \pi_{X}\otimes\pi_{Y}\otimes\pi_{Z}\right)
\circ\left(  \operatorname*{id}\otimes\Delta\right)  \circ\Delta\right)
=X\wedge\left(  Y\wedge Z\right)  .
\]
Comparing these two equalities, we get $\left(  X\wedge Y\right)  \wedge
Z=X\wedge\left(  Y\otimes Z\right)  $ since $\left(  \Delta\otimes
\operatorname*{id}\right)  \circ\Delta=\left(  \operatorname*{id}\otimes
\Delta\right)  \circ\Delta$. This proves Proposition I.3.12 1).

\item \textbf{I.3.4:} In the definition of $N\wedge X$ (shortly before
Proposition I.3.13), you write:%
\[
N\wedge X=\left\{  x\in M,\ \Phi x\in N\otimes C+C\otimes X\right\}  .
\]
The $C\otimes X$ should be $M\otimes X$ here.

\item \textbf{Proposition I.3.13:} Replace $\wedge^{n}M$ by $\wedge^{n}R$ here.

\item \textbf{I.5:} In the definition of a bialgebra, you need one more
condition: the condition that $\varepsilon\left(  1\right)  =1\ \ \ \ $%
\footnote{Theoretically you could also want to exclude the $\varepsilon\left(
1\right)  =1$ condition from the definition of a bialgebra. The only
difference it would make is that it would cause the zero space $0$ to be a
bialgebra (and a Hopf algebra). But I think it is not common to consider $0$
as a bialgebra (and you actually use the condition $\varepsilon\left(
1\right)  =1$ in the proof of Proposition I.7.1).}. (Normally this would
follow from the condition that $\varepsilon$ is an algebra morphism, but
apparently in your text "algebra morphism" does not mean "unital algebra
morphism", and then it does not follow from this condition. Also it does not
follow from any of the four commutative diagrams on page 21.)

\item \textbf{I.6.1:} In the first paragraph of I.6.1, replace "map from
$kG\times kG$ into $kG$" by "map from $kG\times kG$ to $kG$" (since "into"
sounds like a claim that the map is injective).

\item \textbf{I.6.2:} Add a point after "$\varepsilon\mid_{V}=0$".

\item \textbf{Lemma I.6.2:} First, replace $S\left(  \mathcal{H}\right)
\subset\mathcal{H}$ by $S\left(  J\right)  \subset J$. Besides, the usual
definition of a "Hopf ideal" involves a third condition: $\varepsilon\left(
J\right)  =0$. However, this condition is redundant in almost every case - the
only exception is when $J=\mathcal{H}$, which brings us back to the question
whether $0$ should be considered a Hopf algebra.

\item \textbf{End of proof of Proposition I.7.1:} On page 25, replace
$\mathcal{L}\left(  \mathcal{H},\mathcal{H}\otimes H\right)  $ by
$\mathcal{L}\left(  \mathcal{H},\mathcal{H}\otimes\mathcal{H}\right)  $ (the
third $H$ should be calligraphic).

\item \textbf{End of proof of Proposition I.7.1:} On page 25, in the
computation%
\begin{align*}
\left(  \left(  \Delta\circ S\right)  \widetilde{\ast}\Delta\right)  \left(
x\right)   &  =\text{[... several lines of computation ...]}\\
&  =u\circ\varepsilon\left(  x\right)  \otimes u\circ\varepsilon\left(
x\right)  ,
\end{align*}
the term $u\circ\varepsilon\left(  x\right)  \otimes u\circ\varepsilon\left(
x\right)  $ should be replaced by $u\circ\varepsilon\left(  x_{1}\right)
\otimes u\circ\varepsilon\left(  x_{2}\right)  $ (otherwise it wouldn't be
linear in $x$).

\item \textbf{End of proof of Proposition I.7.1:} On page 26, in the
computation%
\begin{align*}
\left(  \Delta\widetilde{\ast}\left(  \tau\circ\left(  S\otimes S\right)
\circ\Delta\right)  \right)  \left(  x\right)   &  =\text{[... several lines
of computation ...]}\\
&  =\left(  \sum_{\left(  x\right)  }x_{1}Sx_{2}\right)  \otimes\left(
u\circ\varepsilon\right)  \left(  x\right) \\
&  =u\circ\varepsilon\left(  x\right)  \otimes u\circ\varepsilon\left(
x\right)  ,
\end{align*}
the last two lines of this computation are incorrect (the terms should be
linear in $x$). I would replace them by%
\begin{align*}
&  =\sum_{\left(  x\right)  }x_{1}\varepsilon\left(  x_{2}\right)
Sx_{3}\otimes1=\left(  \sum_{\left(  x\right)  }x_{1}Sx_{2}\right)
\otimes1=u\circ\varepsilon\left(  x\right)  \otimes1\\
&  =u\circ\varepsilon\left(  x_{1}\right)  \otimes u\circ\varepsilon\left(
x_{2}\right)  ,
\end{align*}


\item \textbf{Proof of Proposition I.7.3:} Replace $S\left(  x\right)  -x$ by
$S\left(  x\right)  +x$ here.

\item \textbf{Proof of Proposition I.7.3:} Replace $\mathbf{1}\otimes\left(
xy+yx\right)  $ by $\mathbf{1}\otimes\left(  xy-yx\right)  $ here.

\item \textbf{II.1:} On page 27, I think it is worth a mention that every
graded bialgebra automatically satisfies $1\in\mathcal{H}_{0}$ and
$\varepsilon\left(  \mathcal{H}_{n}\right)  =0$ for every $n>0$. (More
strongly, every graded counital coalgebra $C$ automatically satisfies
$\varepsilon\left(  C_{n}\right)  =0$ for every $n>0$, and every graded unital
algebra $A$ automatically satisfies $1\in A_{0}$.) This is used when you say
that $\operatorname*{Ker}\varepsilon=\bigoplus\limits_{n\geq1}\mathcal{H}_{n}$
for any connected graded bialgebra $\mathcal{H}$.

\item \textbf{Proposition II.1.1:} Replace $\mathcal{H}^{n}$ by $\mathcal{H}%
_{n}$.

\item \textbf{Proof of Proposition II.1.1:} At the beginning of this proof,
you write:\newline"Thanks to connectedness we clearly can write :%
\[
\Delta x=a\left(  x\otimes1\right)  +b\left(  1\otimes x\right)
+\widetilde{\Delta}x
\]
with $a,b\in k$ and $\widetilde{\Delta}x\in\operatorname*{Ker}\varepsilon
\otimes\operatorname*{Ker}\varepsilon$. The co-unity property then tells us
that, with $k\otimes\mathcal{H}$ and $\mathcal{H}\otimes k$ canonically
identified with $\mathcal{H}$ :%
\begin{align*}
x  &  =\left(  \varepsilon\otimes I\right)  \left(  \Delta x\right)  =bx\\
x  &  =\left(  I\otimes\varepsilon\right)  \left(  \Delta x\right)  =ax,
\end{align*}
hence $a=b=1$."\newline This whole paragraph is slightly flawed. There is a
typo ("co-unity" should be "co-unit"), but I am talking about something more
serious: Connectedness does not \textit{directly} give us $\Delta x=a\left(
x\otimes1\right)  +b\left(  1\otimes x\right)  +\widetilde{\Delta}x$ for some
$a,b\in k$, but rather gives us $\Delta x=u\otimes1+1\otimes
v+\widetilde{\Delta}x$ for some $u,v\in\mathcal{H}_{n}$. We do not yet know
that $u$ and $v$ are multiples of $x$; to see that, we need the counit
property. Hence I would rewrite the above paragraph as follows:\newline"Thanks
to connectedness we clearly can write :%
\[
\Delta x=u\otimes1+1\otimes v+\widetilde{\Delta}x
\]
with $u,v\in\mathcal{H}_{n}$ and $\widetilde{\Delta}x\in\operatorname*{Ker}%
\varepsilon\otimes\operatorname*{Ker}\varepsilon$. The co-unit property then
tells us that, with $k\otimes\mathcal{H}$ and $\mathcal{H}\otimes k$
canonically identified with $\mathcal{H}$ :%
\begin{align*}
x  &  =\left(  \varepsilon\otimes I\right)  \left(  \Delta x\right)  =v\\
x  &  =\left(  I\otimes\varepsilon\right)  \left(  \Delta x\right)  =u,
\end{align*}
hence $\Delta x=x\otimes1+1\otimes x+\widetilde{\Delta}x$."

\item \textbf{Proof of Proposition II.1.1:} Add a point between "$\left\vert
x^{\prime}\right\vert +\left\vert x^{\prime\prime}\right\vert =n$" and "We
easily compute :".

\item \textbf{II.2:} On page 29, in the definition of a "filtered Hopf
algebra", I don't understand the meaning of the word "characteristic" in "a
unit characteristic $u:k\rightarrow\mathcal{H}$".

\item \textbf{II.2:} On page 29, replace "if $x$ is an homogeneous element" by
"if $x$ is a nonzero homogeneous element".

\item \textbf{Proposition II.2.1:} Replace $x\in\mathcal{H}^{n}$ by
$x\in\mathcal{H}^{n}\cap\operatorname*{Ker}\varepsilon$. (Otherwise, $x=1$ and
$n=1$ is a counterexample.)

\item \textbf{Proof of Proposition II.2.1:} I am not sure whether this proof
is really a "Straightforward adaptation of proof of proposition II.1.1". For
example, you cannot apply the co-unit property as easily as you did in the
proof of Proposition II.1.1, since you don't have $\varepsilon\left(
\mathcal{H}^{n}\right)  =0$ for $n\geq1$.\newline Here is what I think is a
correct proof of Proposition II.2.1:\newline First, it is easy to see that any
grouplike element of a filtered coalgebra must lie in the $0$-th part of the
filtration.\footnote{\textit{Proof.} Let $C$ be a filtered coalgebra with
filtration $\left(  C_{n}\right)  _{n\geq0}$. Let $g\in C$ be a grouplike
element of $C$. We must prove that $g\in C_{0}$.
\par
Since $g\in C$, there exists some $n\geq0$ such that $g\in C_{n}$. Let $m$ be
the smallest such $n$. Then, $g\in C_{m}$, but $g\notin C_{m-1}$, where we set
$C_{-1}=0$.
\par
If $m=0$, then we are done, so let us assume that $m>0$. Since $g\in C_{m}$,
we have%
\begin{align*}
\Delta\left(  g\right)   &  \in\Delta\left(  C_{m}\right)  \subseteq
\sum\limits_{i=0}^{m}C_{i}\otimes C_{m-i}=\sum\limits_{i=0}^{m-1}%
\underbrace{C_{i}}_{\subseteq C_{m-1}}\otimes\underbrace{C_{m-i}}_{\subseteq
C_{m}}+C_{m}\otimes\underbrace{C_{0}}_{\subseteq C_{m-1}}\\
&  \subseteq\sum\limits_{i=0}^{m-1}C_{m-1}\otimes C_{m}+C_{m}\otimes
C_{m-1}\subseteq C_{m-1}\otimes C_{m}+C_{m}\otimes C_{m-1}%
\end{align*}
(since $C_{m-1}\otimes C_{m}$ is a $k$-vector space). If $\pi$ denotes the
canonical projection $C_{m}\rightarrow C_{m}\diagup C_{m-1}$, we thus have%
\[
\left(  \pi\otimes\pi\right)  \left(  \Delta\left(  g\right)  \right)
\in\left(  \pi\otimes\pi\right)  \left(  C_{m-1}\otimes C_{m}+C_{m}\otimes
C_{m-1}\right)  \subseteq\underbrace{\pi\left(  C_{m-1}\right)  }_{=0}%
\otimes\pi\left(  C_{m}\right)  +\pi\left(  C_{m}\right)  \otimes
\underbrace{\pi\left(  C_{m-1}\right)  }_{=0}=0,
\]
so that $\left(  \pi\otimes\pi\right)  \left(  \Delta\left(  g\right)
\right)  =0$. But since $\Delta\left(  g\right)  =g\otimes g$ (since $g$ is
grouplike), we have $\left(  \pi\otimes\pi\right)  \left(  \Delta\left(
g\right)  \right)  =\left(  \pi\otimes\pi\right)  \left(  g\otimes g\right)
=\pi\left(  g\right)  \otimes\pi\left(  g\right)  \neq0$ (since $\pi\left(
g\right)  \neq0$, which is because $g\notin C_{m-1}$), and thus we get a
contradiction to $\left(  \pi\otimes\pi\right)  \left(  \Delta\left(
g\right)  \right)  =0$. This contradiction shows that the case $m>0$ cannot
occur. Thus, $m=0$, so that $g\in C_{m}=C_{0}$, qed.} Applied to the grouplike
element $1$ of the filtered coalgebra $\mathcal{H}$, we get $1\in
\mathcal{H}^{0}$. Note that we have not used the connectedness of
$\mathcal{H}$ yet.\newline Next, define a subspace $\mathcal{H}^{+i}$ of
$\mathcal{H}^{i}$ by $\mathcal{H}^{+i}=\mathcal{H}^{i}\cap\operatorname*{Ker}%
\varepsilon$ for every $i>0$. It is easy to see that $\mathcal{H}%
^{i}=\mathcal{H}^{0}+\mathcal{H}^{+i}$ for every $i>0$ (because every
$x\in\mathcal{H}^{i}$ satisfies $x=\underbrace{\varepsilon\left(  x\right)
\cdot1}_{\in\mathcal{H}^{0}}+\underbrace{\left(  x-\varepsilon\left(
x\right)  \cdot1\right)  }_{\in\mathcal{H}^{i}\cap\operatorname*{Ker}%
\varepsilon=\mathcal{H}^{+i}}\in\mathcal{H}^{0}+\mathcal{H}^{+i}$). Now every
$n\geq1$ satisfies%
\[
\Delta\left(  \mathcal{H}^{n}\right)  \subseteq\sum_{p+q=n}\mathcal{H}%
^{p}\otimes\mathcal{H}^{q}=\mathcal{H}^{0}\otimes\mathcal{H}^{n}%
+\sum_{\substack{p+q=n;\\p\neq0;\ q\neq0}}\mathcal{H}^{p}\otimes
\mathcal{H}^{q}+\mathcal{H}^{n}\otimes\mathcal{H}^{0}.
\]
Since every $p\neq0$ and $q\neq0$ with $p+q=n$ satisfy%
\begin{align*}
\underbrace{\mathcal{H}^{p}}_{=\mathcal{H}^{0}+\mathcal{H}^{+p}}%
\otimes\underbrace{\mathcal{H}^{q}}_{=\mathcal{H}^{0}+\mathcal{H}^{+q}}  &
=\left(  \mathcal{H}^{0}+\mathcal{H}^{+p}\right)  \otimes\left(
\mathcal{H}^{0}+\mathcal{H}^{+q}\right) \\
&  =\mathcal{H}^{0}\otimes\underbrace{\mathcal{H}^{0}}_{\subseteq
\mathcal{H}^{n}}+\mathcal{H}^{0}\otimes\underbrace{\mathcal{H}^{+q}%
}_{\subseteq\mathcal{H}^{q}\subseteq\mathcal{H}^{n}}+\underbrace{\mathcal{H}%
^{+p}}_{\subseteq\mathcal{H}^{p}\subseteq\mathcal{H}^{n}}\otimes
\mathcal{H}^{0}+\mathcal{H}^{+p}\otimes\mathcal{H}^{+q}\\
&  \subseteq\mathcal{H}^{0}\otimes\mathcal{H}^{n}+\mathcal{H}^{0}%
\otimes\mathcal{H}^{n}+\mathcal{H}^{n}\otimes\mathcal{H}^{0}+\mathcal{H}%
^{+p}\otimes\mathcal{H}^{+q},
\end{align*}
this becomes%
\begin{align*}
\Delta\left(  \mathcal{H}^{n}\right)   &  \subseteq\mathcal{H}^{0}%
\otimes\mathcal{H}^{n}+\sum_{\substack{p+q=n;\\p\neq0;\ q\neq0}}\left(
\mathcal{H}^{0}\otimes\mathcal{H}^{n}+\mathcal{H}^{0}\otimes\mathcal{H}%
^{n}+\mathcal{H}^{n}\otimes\mathcal{H}^{0}+\mathcal{H}^{+p}\otimes
\mathcal{H}^{+q}\right)  +\mathcal{H}^{n}\otimes\mathcal{H}^{0}\\
&  \subseteq\mathcal{H}^{n}\otimes\mathcal{H}^{0}+\mathcal{H}^{0}%
\otimes\mathcal{H}^{n}+\sum_{\substack{p+q=n;\\p\neq0;\ q\neq0}}\mathcal{H}%
^{+p}\otimes\mathcal{H}^{+q}.
\end{align*}
Thus, for every $n\geq1$ and every $x\in\mathcal{H}^{n}\cap\operatorname*{Ker}%
\varepsilon$, we can write%
\[
\Delta x=u\otimes1+1\otimes v+\widetilde{\Delta}x
\]
for some $u,v\in\mathcal{H}^{n}$ and $\widetilde{\Delta}x\in\sum
\limits_{\substack{p+q=n;\\p\neq0;\ q\neq0}}\mathcal{H}^{+p}\otimes
\mathcal{H}^{+q}$ (here we are using, for the first time, that $\mathcal{H}$
is connected). The co-unit property now tells us that, with $k\otimes
\mathcal{H}$ and $\mathcal{H}\otimes k$ canonically identified with
$\mathcal{H}$, we have%
\begin{align*}
x  &  =\left(  \varepsilon\otimes I\right)  \left(  \Delta x\right)
=v+\varepsilon\left(  u\right)  \ \ \ \ \ \ \ \ \ \ \text{and}\\
x  &  =\left(  I\otimes\varepsilon\right)  \left(  \Delta x\right)
=u+\varepsilon\left(  v\right)
\end{align*}
(here we are using that $\widetilde{\Delta}x\in\sum
\limits_{\substack{p+q=n;\\p\neq0;\ q\neq0}}\mathcal{H}^{+p}\otimes
\mathcal{H}^{+q}$, so that $\left(  \varepsilon\otimes I\right)  \left(
\widetilde{\Delta}x\right)  =0$ and $\left(  I\otimes\varepsilon\right)
\left(  \widetilde{\Delta}x\right)  =0$), and therefore $u=x-\varepsilon
\left(  v\right)  $ and $v=x-\varepsilon\left(  u\right)  $. Hence, $\Delta
x=u\otimes1+1\otimes v+\widetilde{\Delta}x$ rewrites as $\Delta x=x\otimes
1+1\otimes x-\left(  \varepsilon\left(  u\right)  +\varepsilon\left(
v\right)  \right)  1\otimes1+\widetilde{\Delta}x$. Applying $\varepsilon
\otimes I$ to this equation, we get $x=\varepsilon\left(  x\right)
1+x-\left(  \varepsilon\left(  u\right)  +\varepsilon\left(  v\right)
\right)  1$ (since $\left(  \varepsilon\otimes I\right)  \left(
\widetilde{\Delta}x\right)  =0$), which rewrites as $\varepsilon\left(
x\right)  =\varepsilon\left(  u\right)  +\varepsilon\left(  v\right)  $.
Hence, if $x\in\mathcal{H}^{+n}$, we have $\varepsilon\left(  u\right)
+\varepsilon\left(  v\right)  =\varepsilon\left(  x\right)  =0$, so that
$\Delta x=x\otimes1+1\otimes x-\left(  \varepsilon\left(  u\right)
+\varepsilon\left(  v\right)  \right)  1\otimes1+\widetilde{\Delta}x$
simplifies to $\Delta x=x\otimes1+1\otimes x+\widetilde{\Delta}x$.\newline We
thus have proven that every $x\in\mathcal{H}^{+n}$ for every $n>0$ satisfies
$\Delta x=x\otimes1+1\otimes x+\widetilde{\Delta}x$ with $\widetilde{\Delta
}x\in\sum\limits_{\substack{p+q=n;\\p\neq0;\ q\neq0}}\mathcal{H}^{+p}%
\otimes\mathcal{H}^{+q}$. The rest of the proof now indeed proceeds
analogously to the proof of Proposition II.1.1 (except that we don't use homogeneity).

\item \textbf{Proof of Theorem II.2.2:} In the first line of this proof,
replace $S\left(  H^{n}\right)  \subset H^{n}$ by $S\left(  \mathcal{H}%
^{n}\right)  \subset\mathcal{H}^{n}$ (with calligraphic $\mathcal{H}$).

\item \textbf{Proof of Theorem II.2.2:} In the second paragraph of this proof,
replace "inclusion $S\mathcal{H}_{0}\subset\mathcal{H}_{0}$" by "inclusion
$S\mathcal{H}^{0}\subset\mathcal{H}^{0}$".

\item \textbf{Proof of Theorem II.2.2:} In the second paragraph of this proof,
you write%
\[
\mathcal{H}^{n}=\mathcal{H}^{0}\wedge\mathcal{H}^{n-1}=\mathcal{H}^{n-1}\wedge
H^{0}.
\]
The last $H$ should be calligraphic here.

\item \textbf{Proof of Theorem II.2.2:} In the second paragraph of this proof,
replace the formula%
\[
Sx=\sum_{\left(  x\right)  }Sx_{2}\otimes Sx_{1}%
\]
by%
\[
\Delta\left(  Sx\right)  =\sum_{\left(  x\right)  }Sx_{2}\otimes Sx_{1}.
\]


\item \textbf{Proof of Theorem II.2.2:} In the second paragraph of this proof,
replace "its is obviously" by "it is obviously".

\item \textbf{Remark 2 after the proof of Theorem II.2.2:} Replace
"subcoagebra" by "subcoalgebra".

\item \textbf{Proof of Proposition II.3.1:} You write:%
\[
\left(  e-\varphi\right)  ^{\ast k}\left(  x\right)  =m_{\mathcal{A}%
,k-1}\left(  \varphi\otimes\cdots\otimes\varphi\right)  \widetilde{\Delta
}_{k-1}\left(  x\right)  .
\]
This should be%
\begin{align*}
\left(  e-\varphi\right)  ^{\ast k}\left(  x\right)   &  =m_{\mathcal{A}%
,k-1}\left(  \left(  e-\varphi\right)  \otimes\cdots\otimes\left(
e-\varphi\right)  \right)  \Delta_{k-1}\left(  x\right) \\
&  =m_{\mathcal{A},k-1}\left(  \left(  -\varphi\right)  \otimes\cdots
\otimes\left(  -\varphi\right)  \right)  \widetilde{\Delta}_{k-1}\left(
x\right)  .
\end{align*}


\item \textbf{II.3:} Between the proof of Corollary II.3.2 and Proposition
II.3.3, you write: "For any $x\in\mathcal{H}^{n}$ the exponential :%
\[
e^{\ast\alpha}\left(  x\right)  =\sum_{k\geq0}\dfrac{\alpha^{\ast k}\left(
x\right)  }{k!}%
\]
is a finite sum (ending up at $k=n$)." You could add "and any $\alpha
\in\mathfrak{g}$" after "For any $x\in\mathcal{H}^{n}$" here, in order to make
it clear what $\alpha$ is.

\item \textbf{Proposition II.3.3:} In part 2), replace "then" by "the".

\item \textbf{II.3:} On page 32, in the definition of $\mathcal{L}_{n}$, you
falsely write $\mathcal{L}^{n}$ instead of $\mathcal{L}_{n}$. The same mistake
is repeated on page 33, in the last line of II.3 ("the Lie algebras
$\mathfrak{g}/\mathcal{L}^{n}$"). Also the same mistake, this time is repeated
in Proposition II.3 (this time the $\mathcal{L}^{p}$, $\mathcal{L}^{q}$ and
$\mathcal{L}^{p+q}$ should be $\mathcal{L}_{p}$, $\mathcal{L}_{q}$ and
$\mathcal{L}_{p+q}$).

\item \textbf{II.4:} When you define "characters", replace "algebra morphisms"
by "unital algebra morphisms" (otherwise, $0$ would be a character,
contradicting Proposition II.4.1 3)).

\item \textbf{Proof of Proposition II.4.1:} In the commutative diagram on page
34, replace $\otimes\mathcal{A}$ by $\mathcal{A}\otimes\mathcal{A}$.

\item \textbf{Proof of Proposition II.4.1:} This proof ends with the equation%
\[
\tau^{-1}\left(  x\right)  =\sum_{k\geq0}\left(  e-\tau\right)  ^{\ast
k}\left(  x\right)  .
\]
To keep notations consistent, I believe you should replace $\tau^{-1}$ by
$\tau^{\ast-1}$ here.

\item A remark about Proposition II.4.1: The following statement generalizes
Proposition II.4.1 2):\newline If $\mathcal{H}$ is a Hopf algebra over a field
$k$ (not necessarily of characteristic $0$), and $\mathcal{A}$ is a
$k$-algebra, and if $\xi:\mathcal{H}\rightarrow\mathcal{A}$ is a cocycle which
has an inverse with respect to the convolution, then this inverse $\xi
^{\ast-1}$ is a cocycle as well.\newline\textit{Proof of this statement:}
Define a $k$-linear map $\Phi_{1}:\mathcal{H}\otimes\mathcal{H}\rightarrow
\mathcal{A}$ by%
\[
\Phi_{1}\left(  x\otimes y\right)  =\xi\left(  xy\right)
\ \ \ \ \ \ \ \ \ \ \text{for every }x\in\mathcal{H}\text{ and }%
y\in\mathcal{H}\text{.}%
\]
\newline Define a $k$-linear map $\Phi_{2}:\mathcal{H}\otimes\mathcal{H}%
\rightarrow\mathcal{A}$ by%
\[
\Phi_{2}\left(  x\otimes y\right)  =\xi^{\ast-1}\left(  xy\right)
\ \ \ \ \ \ \ \ \ \ \text{for every }x\in\mathcal{H}\text{ and }%
y\in\mathcal{H}\text{.}%
\]
\newline Define a $k$-linear map $\Phi_{3}:\mathcal{H}\otimes\mathcal{H}%
\rightarrow\mathcal{A}$ by%
\[
\Phi_{3}\left(  x\otimes y\right)  =\xi^{\ast-1}\left(  yx\right)
\ \ \ \ \ \ \ \ \ \ \text{for every }x\in\mathcal{H}\text{ and }%
y\in\mathcal{H}\text{.}%
\]
\newline Then, every $x\in\mathcal{H}$ and $y\in\mathcal{H}$ satisfy%
\begin{align*}
\left(  \Phi_{1}\ast\Phi_{2}\right)  \left(  x\otimes y\right)   &
=\sum_{\left(  x\otimes y\right)  }\Phi_{1}\left(  \left(  x\otimes y\right)
_{1}\right)  \Phi_{2}\left(  \left(  x\otimes y\right)  _{2}\right)
=\sum_{\left(  x\right)  \left(  y\right)  }\underbrace{\Phi_{1}\left(
x_{1}\otimes y_{1}\right)  }_{=\xi\left(  x_{1}y_{1}\right)  }\underbrace{\Phi
_{2}\left(  x_{2}\otimes y_{2}\right)  }_{=\xi^{\ast-1}\left(  x_{2}%
y_{2}\right)  }\\
&  =\sum_{\left(  x\right)  \left(  y\right)  }\xi\left(  x_{1}y_{1}\right)
\xi^{\ast-1}\left(  x_{2}y_{2}\right)  =\sum_{\left(  xy\right)  }\xi\left(
\left(  xy\right)  _{1}\right)  \xi^{\ast-1}\left(  \left(  xy\right)
_{2}\right)  \\
&  =\underbrace{\left(  \xi\ast\xi^{\ast-1}\right)  }_{=e}\left(  xy\right)
=e\left(  xy\right)  =e\left(  x\otimes y\right)  ,
\end{align*}
so that $\Phi_{1}\ast\Phi_{2}=e$. Also, every $x\in\mathcal{H}$ and
$y\in\mathcal{H}$ satisfy%
\begin{align*}
\left(  \Phi_{3}\ast\Phi_{1}\right)  \left(  x\otimes y\right)   &
=\sum_{\left(  x\otimes y\right)  }\Phi_{3}\left(  \left(  x\otimes y\right)
_{1}\right)  \Phi_{1}\left(  \left(  x\otimes y\right)  _{2}\right)
=\sum_{\left(  x\right)  \left(  y\right)  }\underbrace{\Phi_{3}\left(
x_{1}\otimes y_{1}\right)  }_{=\xi^{\ast-1}\left(  y_{1}x_{1}\right)
}\underbrace{\Phi_{1}\left(  x_{2}\otimes y_{2}\right)  }_{\substack{=\xi
\left(  x_{2}y_{2}\right)  =\xi\left(  y_{2}x_{2}\right)  \\\text{(since }%
\xi\text{ is a cocycle)}}}\\
&  =\sum_{\left(  x\right)  \left(  y\right)  }\xi^{\ast-1}\left(  y_{1}%
x_{1}\right)  \xi\left(  y_{2}x_{2}\right)  =\sum_{\left(  yx\right)  }%
\xi^{\ast-1}\left(  \left(  yx\right)  _{1}\right)  \xi\left(  \left(
yx\right)  _{2}\right)  \\
&  =\underbrace{\left(  \xi^{\ast-1}\ast\xi\right)  }_{=e}\left(  yx\right)
=e\left(  yx\right)  =e\left(  y\right)  e\left(  x\right)  =e\left(  x\otimes
y\right)  ,
\end{align*}
so that $\Phi_{3}\ast\Phi_{1}=e$. Thus, $\Phi_{2}=\underbrace{e}_{=\Phi
_{3}\ast\Phi_{1}}\ast\Phi_{2}=\Phi_{3}\ast\underbrace{\Phi_{1}\ast\Phi_{2}%
}_{=e}=\Phi_{3}\ast e=\Phi_{3}$. Thus, every $x\in\mathcal{H}$ and
$y\in\mathcal{H}$ satisfy $\xi^{\ast-1}\left(  xy\right)  =\underbrace{\Phi
_{2}}_{=\Phi_{3}}\left(  x\otimes y\right)  =\Phi_{3}\left(  x\otimes
y\right)  =\xi^{\ast-1}\left(  yx\right)  $. In other words, $\xi^{\ast-1}$ is
a cocycle, qed.

\item \textbf{Proof of Proposition II.4.2:} Replace $\sum\limits_{\left(
x\right)  \left(  y\right)  }\alpha\left(  x_{1}x_{2}\right)  \beta\left(
y_{1}y_{2}\right)  $ by $\sum\limits_{\left(  x\right)  \left(  y\right)
}\alpha\left(  x_{1}y_{1}\right)  \beta\left(  x_{2}y_{2}\right)  $. Also, one
line further below, replace $e\left(  x_{2}\right)  \alpha\left(
y_{2}\right)  $ by $e\left(  x_{2}\right)  \beta\left(  y_{2}\right)  $.

\item \textbf{Proof of Proposition II.4.2:} I fear you don't really prove that
the exponential restricts to a bijection from $\mathfrak{g}_{1}$ onto $G_{1}$;
instead you only show that it maps $\mathfrak{g}_{1}$ into $G_{1}$ (but not
necessarily surjectively). Do you have an easy proof for the fact that it
restricts to a bijection from $\mathfrak{g}_{1}$ onto $G_{1}$ ? Here is the
only proof I have:\newline\textit{Proof.} You have shown that the exponential
maps $\mathfrak{g}_{1}$ into $G_{1}$. Now it remains to show that any
$\alpha\in\mathfrak{g}$ satisfying $e^{\ast\alpha}\in G_{1}$ must lie in
$\mathfrak{g}_{1}$.\newline So consider some $\alpha\in\mathfrak{g}$
satisfying $e^{\ast\alpha}\in G_{1}$.\newline Any two elements $\beta$ and
$\gamma$ of $\mathfrak{g}$ which commute satisfy $e^{\ast\left(  \beta
+\gamma\right)  }=e^{\ast\beta}\cdot e^{\ast\gamma}$.\ \ \ \ \footnote{This
can be proven by the same argument as the classical one used to prove that
$\exp\left(  x+y\right)  =\exp x\cdot\exp y$ for two reals $x$ and $y$ (where
$\exp$ is defined by the power series).} Using this fact and induction over
$n$, we can prove the following: Every $m\in\mathbb{N}$ satisfies $e^{\ast
m\alpha}=\left(  e^{\ast\alpha}\right)  ^{m}$. Thus, $e^{\ast m\alpha}\in
G_{1}$ for every $m\in\mathbb{N}$ (since $e^{\ast\alpha}\in G_{1}$ and since
$G_{1}$ is a group).\newline Now let $x\in\mathcal{H}$ and $y\in\mathcal{H}$
be arbitrary. We will prove that $\alpha\left(  xy\right)  =e\left(  x\right)
\alpha\left(  y\right)  +\alpha\left(  x\right)  e\left(  y\right)  $.\newline
Since $x\in\mathcal{H}$, there exists some $i\in\mathbb{N}$ such that
$x\in\mathcal{H}^{i}$. Consider this $i$. Then, the power series $e^{\ast
m\alpha}\left(  x\right)  =\sum\limits_{k\geq0}\dfrac{\left(  m\alpha\right)
^{\ast k}\left(  x\right)  }{k!}$ ends up at $k=i$.\newline Since
$y\in\mathcal{H}$, there exists some $j\in\mathbb{N}$ such that $y\in
\mathcal{H}^{j}$. Consider this $j$. Then, the power series $e^{\ast m\alpha
}\left(  y\right)  =\sum\limits_{k\geq0}\dfrac{\left(  m\alpha\right)  ^{\ast
k}\left(  y\right)  }{k!}$ ends up at $k=j$.\newline Also, $\underbrace{x}%
_{\in\mathcal{H}^{i}}\underbrace{y}_{\in\mathcal{H}^{j}}\in\mathcal{H}%
^{i}\mathcal{H}^{j}\subseteq\mathcal{H}^{i+j}$. Thus, the power series
$e^{\ast m\alpha}\left(  y\right)  =\sum\limits_{k\geq0}\dfrac{\left(
m\alpha\right)  ^{\ast k}\left(  y\right)  }{k!}$ ends up at $k=i+j$.\newline
Every $m\in\mathbb{N}$ satisfies%
\begin{align*}
e^{\ast m\alpha}\left(  xy\right)   &  =e^{\ast m\alpha}\left(  x\right)
e^{\ast m\alpha}\left(  y\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{since
}e^{\ast m\alpha}\in G_{1}\text{, so that }e^{\ast m\alpha}\text{ is a
character}\right) \\
&  =\sum\limits_{k\geq0}\dfrac{\left(  m\alpha\right)  ^{\ast k}\left(
x\right)  }{k!}\sum\limits_{k\geq0}\dfrac{\left(  m\alpha\right)  ^{\ast
k}\left(  y\right)  }{k!}\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition
of }e^{\ast m\alpha}\right) \\
&  =\sum\limits_{k\geq0,\ \ell\geq0}\underbrace{\dfrac{\left(  m\alpha\right)
^{\ast k}\left(  x\right)  \cdot\left(  m\alpha\right)  ^{\ast\ell}\left(
y\right)  }{k!\ell!}}_{=m^{k+\ell}\dfrac{\alpha^{\ast k}\left(  x\right)
\cdot\alpha^{\ast\ell}\left(  y\right)  }{k!\ell!}}=\sum\limits_{k\geq
0,\ \ell\geq0}m^{k+\ell}\dfrac{\alpha^{\ast k}\left(  x\right)  \cdot
\alpha^{\ast\ell}\left(  y\right)  }{k!\ell!}\\
&  =\sum_{n\geq0}m^{n}\sum_{k+\ell=n}\dfrac{\alpha^{\ast k}\left(  x\right)
\cdot\alpha^{\ast\ell}\left(  y\right)  }{k!\ell!}%
\end{align*}
and thus%
\begin{align}
\sum_{n\geq0}m^{n}\sum_{k+\ell=n}\dfrac{\alpha^{\ast k}\left(  x\right)
\cdot\alpha^{\ast\ell}\left(  y\right)  }{k!\ell!}  &  =e^{\ast m\alpha
}\left(  xy\right)  =\sum\limits_{k\geq0}\underbrace{\dfrac{\left(
m\alpha\right)  ^{\ast k}\left(  xy\right)  }{k!}}_{=m^{k}\dfrac{\alpha^{\ast
k}\left(  xy\right)  }{k!}}\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition
of }e^{\ast m\alpha}\right) \nonumber\\
&  =\sum\limits_{k\geq0}m^{k}\dfrac{\alpha^{\ast k}\left(  xy\right)  }%
{k!}=\sum_{n\geq0}m^{n}\dfrac{\alpha^{\ast n}\left(  xy\right)  }{n!}.
\label{polyID}%
\end{align}
Thus, the identity (\ref{polyID}) holds for infinitely many distinct values of
$m\in\mathcal{A}$ (because it holds for every $m\in\mathbb{N}$, and because
$\mathbb{N}$ injects into $\mathcal{A}$\ \ \ \ \footnote{Here we have used the
condition that $\operatorname*{char}k=0$.}). But (\ref{polyID}) is a
polynomial identity in $m$, since both sums $\sum\limits_{n\geq0}m^{n}%
\sum\limits_{k+\ell=n}\dfrac{\alpha^{\ast k}\left(  x\right)  \cdot
\alpha^{\ast\ell}\left(  y\right)  }{k!\ell!}$ and $\sum\limits_{n\geq0}%
m^{n}\dfrac{\alpha^{\ast n}\left(  xy\right)  }{n!}$ end up at $n=i+j$ (since
every $n>i+j$ satisfies $\sum\limits_{k+\ell=n}\dfrac{\alpha^{\ast k}\left(
x\right)  \cdot\alpha^{\ast\ell}\left(  y\right)  }{k!\ell!}=0$%
\ \ \ \ \footnote{\textit{Proof.} Let $n>i+j$ be arbitrary. Any $k$ and $\ell$
with $k+\ell=n$ satisfy at least one of the two inequalities $k>i$ and
$\ell>j$ (since otherwise, we would have $k\leq i$ and $\ell\leq j$, so that
$k+\ell\leq i+j<n$, contradicting $k+\ell=n$). But in each of these two cases
we have $\alpha^{\ast k}\left(  x\right)  \cdot\alpha^{\ast\ell}\left(
y\right)  =0$ (in fact, in the case $k>i$ we have $\alpha^{\ast k}\left(
x\right)  =0$ (because $x\in\mathcal{H}^{i}$), whereas in the case $\ell>j$ we
have $\alpha^{\ast\ell}\left(  y\right)  =0$ (because $y\in\mathcal{H}^{j}$)).
Thus, any $k$ and $\ell$ with $k+\ell=n$ satisfy $\alpha^{\ast k}\left(
x\right)  \cdot\alpha^{\ast\ell}\left(  y\right)  =0$. Hence, $\sum
\limits_{k+\ell=n}\dfrac{\alpha^{\ast k}\left(  x\right)  \cdot\alpha
^{\ast\ell}\left(  y\right)  }{k!\ell!}=\sum\limits_{k+\ell=n}\dfrac{0}%
{k!\ell!}=0$.} and $\dfrac{\alpha^{\ast n}\left(  xy\right)  }{n!}=0$ (since
$xy\in\mathcal{H}^{i+j}$)). Since this polynomial identity (\ref{polyID})
holds for infinitely many distinct values of $m\in\mathcal{A}$, it must
therefore hold as a formal polynomial identity, i. e., we must have%
\begin{equation}
\sum_{n\geq0}X^{n}\sum_{k+\ell=n}\dfrac{\alpha^{\ast k}\left(  x\right)
\cdot\alpha^{\ast\ell}\left(  y\right)  }{k!\ell!}=\sum_{n\geq0}X^{n}%
\dfrac{\alpha^{\ast n}\left(  xy\right)  }{n!} \label{polyIDX}%
\end{equation}
as an identity between elements of the polynomial ring $\mathcal{A}\left[
X\right]  $. But if two polynomials are equal as elements of the polynomial
ring $\mathcal{A}\left[  X\right]  $, their corresponding coefficients must be
equal to each other; therefore, we can compare coefficients in (\ref{polyIDX})
and conclude that every $n\geq0$ satisfies $\sum\limits_{k+\ell=n}%
\dfrac{\alpha^{\ast k}\left(  x\right)  \cdot\alpha^{\ast\ell}\left(
y\right)  }{k!\ell!}=\dfrac{\alpha^{\ast n}\left(  xy\right)  }{n!}$. Applying
this to $n=1$, we get $\sum\limits_{k+\ell=1}\dfrac{\alpha^{\ast k}\left(
x\right)  \cdot\alpha^{\ast\ell}\left(  y\right)  }{k!\ell!}=\dfrac
{\alpha^{\ast1}\left(  xy\right)  }{1!}$. Since%
\begin{align*}
\sum\limits_{k+\ell=1}\dfrac{\alpha^{\ast k}\left(  x\right)  \cdot
\alpha^{\ast\ell}\left(  y\right)  }{k!\ell!}  &  =\dfrac{\alpha^{\ast
0}\left(  x\right)  \cdot\alpha^{\ast1}\left(  y\right)  }{0!1!}+\dfrac
{\alpha^{\ast1}\left(  x\right)  \cdot\alpha^{\ast0}\left(  y\right)  }%
{1!0!}=\dfrac{e\left(  x\right)  \cdot\alpha\left(  y\right)  }{1}%
+\dfrac{\alpha\left(  x\right)  \cdot e\left(  y\right)  }{1}\\
&  =e\left(  x\right)  \alpha\left(  y\right)  +\alpha\left(  x\right)
e\left(  y\right)
\end{align*}
and $\dfrac{\alpha^{\ast1}\left(  xy\right)  }{1!}=\alpha\left(  xy\right)  $,
this rewrites as $e\left(  x\right)  \alpha\left(  y\right)  +\alpha\left(
x\right)  e\left(  y\right)  =\alpha\left(  xy\right)  $. Since this holds for
any $x\in\mathcal{H}$ and $y\in\mathcal{H}$, we thus conclude that $\alpha$ is
a derivation. In other words, $\alpha\in\mathfrak{g}$, qed.

\item \textbf{Theorem II.5.1:} Replace "$\operatorname*{Ker}\varepsilon$ into
$\mathcal{A}_{-}$" by "$\operatorname*{Ker}\varepsilon$ to $\mathcal{A}_{-}$".
Also, replace "$\mathcal{H}$ into $A_{+}$" by "$\mathcal{H}$ to $\mathcal{A}%
_{+}$" (this includes replacing the $A$ by a calligraphic $\mathcal{A}$).

\item \textbf{Proof of Theorem II.5.1, part 1):} In the "easy computation" you
make in order to check $\varphi_{+}=\varphi_{-}\ast\varphi$, remove the point
after $\left(  I-\pi\right)  \left(  \varphi\left(  x\right)  +\sum
\limits_{\left(  x\right)  }\varphi_{-}\left(  x^{\prime}\right)
\varphi\left(  x^{\prime\prime}\right)  \right)  $.

\item \textbf{Proof of Theorem II.5.1:} The proof of part 1) of this theorem
is not complete: it is not clear whether the recurrence equation%
\begin{equation}
\varphi_{-}\left(  x\right)  =-\pi\left(  \varphi\left(  x\right)
+\sum_{\left(  x\right)  }\varphi_{-}\left(  x^{\prime}\right)  \varphi\left(
x^{\prime\prime}\right)  \right)  \label{II.5.1.rec}%
\end{equation}
defining the function $\varphi_{-}$ is "stable" in the sense that if we have
take some $x\in\mathcal{H}^{n}$, then we get one and the same value of
$\varphi_{-}\left(  x\right)  $ no matter whether we treat $x$ as an element
of $\mathcal{H}^{n}$ and apply (\ref{II.5.1.rec}) $n$ times or we treat $x$ as
an element of $\mathcal{H}^{n+1}$ and apply (\ref{II.5.1.rec}) $n+1$
times.\newline Here is how I would fix this proof:\newline\textit{Proof of
part 1) of Theorem II.5.1:} For every $n\in\mathbb{N}$, let $\mathcal{H}^{+n}$
be the subspace $\mathcal{H}^{n}\cap\operatorname*{Ker}\varepsilon$ of
$\operatorname*{Ker}\varepsilon$. It is easy to see that $\operatorname*{Ker}%
\varepsilon=\bigcup\limits_{n\geq0}\mathcal{H}^{+n}$ and $\mathcal{H}^{+0}%
=0$.\newline For every $n\in\mathbb{N}$ we will now define two maps
$\varphi_{n-}:\mathcal{H}^{+n}\rightarrow\mathcal{A}$ and $\varphi
_{n+}:\mathcal{H}^{+n}\rightarrow\mathcal{A}$. We do this by induction over
$n$:\newline For $n=0$, define both maps $\varphi_{n-}$ and $\varphi_{n+}$ to
be the zero map (this is the only choice anyway, since $\mathcal{H}^{+0}%
=0$).\newline Let $m\in\mathbb{N}$. Assume that we already have defined two
maps $\varphi_{m-}:\mathcal{H}^{+m}\rightarrow\mathcal{A}$ and $\varphi
_{m+}:\mathcal{H}^{+m}\rightarrow\mathcal{A}$. Then we define two maps
$\varphi_{\left(  m+1\right)  -}:\mathcal{H}^{+\left(  m+1\right)
}\rightarrow\mathcal{A}$ and $\varphi_{\left(  m+1\right)  +}:\mathcal{H}%
^{+\left(  m+1\right)  }\rightarrow\mathcal{A}$ by%
\begin{equation}
\varphi_{\left(  m+1\right)  -}\left(  x\right)  =-\pi\left(  \varphi\left(
x\right)  +\sum_{\left(  x\right)  }\varphi_{m-}\left(  x^{\prime}\right)
\varphi\left(  x^{\prime\prime}\right)  \right)  \ \ \ \ \ \ \ \ \ \ \text{for
every }x\in\mathcal{H}^{+\left(  m+1\right)  } \label{rec1}%
\end{equation}
and%
\begin{equation}
\varphi_{\left(  m+1\right)  +}\left(  x\right)  =\left(  I-\pi\right)
\left(  \varphi\left(  x\right)  +\sum_{\left(  x\right)  }\varphi_{m-}\left(
x^{\prime}\right)  \varphi\left(  x^{\prime\prime}\right)  \right)
\ \ \ \ \ \ \ \ \ \ \text{for every }x\in\mathcal{H}^{+\left(  m+1\right)  },
\label{rec2}%
\end{equation}
respectively.\footnote{This definition rests on the fact that every
$x\in\mathcal{H}^{+\left(  m+1\right)  }$ satisfies $\sum\limits_{\left(
x\right)  }x^{\prime}\otimes x^{\prime\prime}=\widetilde{\Delta}x\in
\sum\limits_{\substack{p+q=m+1;\\p\neq0;\ q\neq0}}\underbrace{\mathcal{H}%
^{+p}}_{\subseteq\mathcal{H}^{+m}}\otimes\underbrace{\mathcal{H}^{+q}%
}_{\subseteq\mathcal{H}}\subseteq\sum\limits_{\substack{p+q=m+1;\\p\neq
0;\ q\neq0}}\mathcal{H}^{+m}\otimes\mathcal{H}\subseteq\mathcal{H}^{+m}%
\otimes\mathcal{H}$.}\newline Now we will prove that every $n\in\mathbb{N}$
satisfies the three equations%
\begin{align}
\varphi_{\left(  n+1\right)  -}\mid_{\mathcal{H}^{+n}}  &  =\varphi
_{n-},\label{ind1}\\
\varphi_{\left(  n+1\right)  +}\mid_{\mathcal{H}^{+n}}  &  =\varphi
_{n+},\label{ind2}\\
\varphi_{n+}  &  =\varphi_{n-}\ast\varphi. \label{ind3}%
\end{align}
(The last of these equations is supposed to mean that every $x\in
\mathcal{H}^{+n}$ satisfies $\varphi_{n+}\left(  x\right)  =\sum
\limits_{\left(  x\right)  }\varphi_{n-}\left(  x_{1}\right)  \varphi\left(
x_{2}\right)  $.)\newline We will do this by induction over $n$: The induction
base (the case $n=0$) is trivial since $\mathcal{H}^{0}=0$. Now to the
induction step: Consider some positive $m\in\mathbb{N}$. Assume that the
equations (\ref{ind1}), (\ref{ind2}), (\ref{ind3}) are all proven for $n=m-1$.
Now let us prove (\ref{ind1}), (\ref{ind2}), (\ref{ind3}) for $n=m$.\newline
Since the equations (\ref{ind1}), (\ref{ind2}), (\ref{ind3}) are all proven
for $n=m-1$, we have $\varphi_{m-}\mid_{\mathcal{H}^{+\left(  m-1\right)  }%
}=\varphi_{\left(  m-1\right)  -}$, $\varphi_{m+}\mid_{\mathcal{H}^{+\left(
m-1\right)  }}=\varphi_{\left(  m-1\right)  +}$ and $\varphi_{\left(
m-1\right)  +}=\varphi_{\left(  m-1\right)  -}\ast\varphi$.\newline Let
$x\in\mathcal{H}^{m}$ be arbitrary. Then,%
\[
\left(  \varphi_{\left(  m+1\right)  -}\mid_{\mathcal{H}^{+m}}\right)  \left(
x\right)  =\varphi_{\left(  m+1\right)  -}\left(  x\right)  =-\pi\left(
\varphi\left(  x\right)  +\sum_{\left(  x\right)  }\varphi_{m-}\left(
x^{\prime}\right)  \varphi\left(  x^{\prime\prime}\right)  \right)  .
\]
Since $x\in\mathcal{H}^{m}$ yields $\sum\limits_{\left(  x\right)  }x^{\prime
}\otimes x^{\prime\prime}\in\mathcal{H}^{+\left(  m-1\right)  }\otimes
\mathcal{H}$\ \ \ \ \footnote{In fact, $\sum\limits_{\left(  x\right)
}x^{\prime}\otimes x^{\prime\prime}=\widetilde{\Delta}x\in\sum
\limits_{\substack{p+q=m;\\p\neq0;\ q\neq0}}\underbrace{\mathcal{H}^{+p}%
}_{\subseteq\mathcal{H}^{+\left(  m-1\right)  }}\otimes\underbrace{\mathcal{H}%
^{+q}}_{\subseteq\mathcal{H}}\subseteq\sum\limits_{\substack{p+q=m;\\p\neq
0;\ q\neq0}}\mathcal{H}^{+\left(  m-1\right)  }\otimes\mathcal{H}%
\subseteq\mathcal{H}^{+\left(  m-1\right)  }\otimes\mathcal{H}$.}, we can
assume that $x^{\prime}\in\mathcal{H}^{+\left(  m-1\right)  }$ in this
equation, and thus we get%
\begin{align*}
\left(  \varphi_{\left(  m+1\right)  -}\mid_{\mathcal{H}^{m}}\right)  \left(
x\right)   &  =-\pi\left(  \varphi\left(  x\right)  +\sum_{\left(  x\right)
}\underbrace{\varphi_{m-}\left(  x^{\prime}\right)  }_{\substack{=\left(
\varphi_{m-}\mid_{\mathcal{H}^{+\left(  m-1\right)  }}\right)  \left(
x^{\prime}\right)  \\\text{(since }x^{\prime}\in\mathcal{H}^{+\left(
m-1\right)  }\text{)}}}\varphi\left(  x^{\prime\prime}\right)  \right) \\
&  =-\pi\left(  \varphi\left(  x\right)  +\sum_{\left(  x\right)
}\underbrace{\left(  \varphi_{m-}\mid_{\mathcal{H}^{+\left(  m-1\right)  }%
}\right)  }_{=\varphi_{\left(  m-1\right)  -}}\left(  x^{\prime}\right)
\varphi\left(  x^{\prime\prime}\right)  \right) \\
&  =-\pi\left(  \varphi\left(  x\right)  +\sum_{\left(  x\right)  }%
\varphi_{\left(  m-1\right)  -}\left(  x^{\prime}\right)  \varphi\left(
x^{\prime\prime}\right)  \right)  .
\end{align*}
But comparing this to%
\begin{align*}
\varphi_{m-}\left(  x\right)   &  =-\pi\left(  \varphi\left(  x\right)
+\sum_{\left(  x\right)  }\varphi_{\left(  m-1\right)  -}\left(  x^{\prime
}\right)  \varphi\left(  x^{\prime\prime}\right)  \right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{rec1}), applied to }m-1\text{
instead of }m\right)  ,
\end{align*}
we obtain $\left(  \varphi_{\left(  m+1\right)  -}\mid_{\mathcal{H}^{+m}%
}\right)  \left(  x\right)  =\varphi_{m-}\left(  x\right)  $. Since this holds
for every $x\in\mathcal{H}^{m}$, we thus have showed that $\varphi_{\left(
m+1\right)  -}\mid_{\mathcal{H}^{+m}}=\varphi_{m-}$. In other words, we proved
(\ref{ind1}) for $n=m$. Similarly to our proof of $\varphi_{\left(
m+1\right)  -}\mid_{\mathcal{H}^{+m}}=\varphi_{m-}$, we can show that
$\varphi_{\left(  m+1\right)  +}\mid_{\mathcal{H}^{+m}}=\varphi_{m+}$. Thus,
we proved (\ref{ind2}) for $n=m$. To complete the induction step, we now need
to verify (\ref{ind3}) for $n=m$.\newline Let $x\in\mathcal{H}^{m}$. Since
$\varphi_{m-}=\varphi_{\left(  m+1\right)  -}\mid_{\mathcal{H}^{+m}}$, we get%
\[
\varphi_{m-}\left(  x\right)  =\left(  \varphi_{\left(  m+1\right)  -}%
\mid_{\mathcal{H}^{+m}}\right)  \left(  x\right)  =\varphi_{\left(
m+1\right)  -}\left(  x\right)  =-\pi\left(  \varphi\left(  x\right)
+\sum_{\left(  x\right)  }\varphi_{m-}\left(  x^{\prime}\right)
\varphi\left(  x^{\prime\prime}\right)  \right)  .
\]
But since $\varphi_{m+}=\varphi_{\left(  m+1\right)  +}\mid_{\mathcal{H}^{+m}%
}$, we have%
\begin{align*}
\varphi_{m+}\left(  x\right)   &  =\left(  \varphi_{\left(  m+1\right)  +}%
\mid_{\mathcal{H}^{+m}}\right)  \left(  x\right)  =\varphi_{\left(
m+1\right)  +}\left(  x\right)  =\left(  I-\pi\right)  \left(  \varphi\left(
x\right)  +\sum_{\left(  x\right)  }\varphi_{m-}\left(  x^{\prime}\right)
\varphi\left(  x^{\prime\prime}\right)  \right) \\
&  =\varphi\left(  x\right)  +\sum_{\left(  x\right)  }\varphi_{m-}\left(
x^{\prime}\right)  \varphi\left(  x^{\prime\prime}\right)  -\pi\left(
\varphi\left(  x\right)  +\sum_{\left(  x\right)  }\varphi_{m-}\left(
x^{\prime}\right)  \varphi\left(  x^{\prime\prime}\right)  \right) \\
&  =\varphi\left(  x\right)  +\sum_{\left(  x\right)  }\varphi_{m-}\left(
x^{\prime}\right)  \varphi\left(  x^{\prime\prime}\right)
+\underbrace{\left(  -\pi\left(  \varphi\left(  x\right)  +\sum_{\left(
x\right)  }\varphi_{m-}\left(  x^{\prime}\right)  \varphi\left(
x^{\prime\prime}\right)  \right)  \right)  }_{=\varphi_{m-}\left(  x\right)
}\\
&  =\varphi\left(  x\right)  +\sum_{\left(  x\right)  }\varphi_{m-}\left(
x^{\prime}\right)  \varphi\left(  x^{\prime\prime}\right)  +\varphi
_{m-}\left(  x\right)  =\sum_{\left(  x\right)  }\varphi_{m-}\left(
x_{1}\right)  \varphi\left(  x_{2}\right)  =\left(  \varphi_{m-}\ast
\varphi\right)  \left(  x\right)  .
\end{align*}
Since this is proven for every $x\in\mathcal{H}^{m}$, we conclude that
$\varphi_{m+}=\varphi_{m-}\ast\varphi$. Thus, (\ref{ind3}) is verified for
$n=m$.\newline We have thus proven the three equations (\ref{ind1}),
(\ref{ind2}), (\ref{ind3}) for $n=m$. This completes the induction step, and
thus we have proven (\ref{ind1}), (\ref{ind2}), (\ref{ind3}) for all
$n\in\mathbb{N}$.\newline Note that $I-\pi$ is a projection onto
$\mathcal{A}_{+}$ (since $\pi$ is a projection parallel to $\mathcal{A}_{+}%
$).\newline For every $x\in\operatorname*{Ker}\varepsilon$, let $N\left(
x\right)  $ be the smallest $m\in\mathbb{N}$ satisfying $x\in\mathcal{H}^{m}%
$.\newline Now, let us define a map $\varphi_{\infty-}:\operatorname*{Ker}%
\varepsilon\rightarrow\mathcal{A}$ by setting
\[
\varphi_{\infty-}\left(  x\right)  =\varphi_{\left(  N\left(  x\right)
\right)  -}\left(  x\right)  \ \ \ \ \ \ \ \ \ \ \text{for every }%
x\in\operatorname*{Ker}\varepsilon.
\]
Then, (\ref{ind1}) shows that%
\[
\varphi_{\infty-}\left(  x\right)  =\varphi_{m-}\left(  x\right)
\ \ \ \ \ \ \ \ \ \ \text{for every }x\in\operatorname*{Ker}\varepsilon\text{
and every }m\in\mathbb{N}\text{ satisfying }x\in\mathcal{H}^{m}.
\]
\newline Similarly, let us define a map $\varphi_{\infty+}:\operatorname*{Ker}%
\varepsilon\rightarrow\mathcal{A}$ by setting
\[
\varphi_{\infty+}\left(  x\right)  =\varphi_{\left(  N\left(  x\right)
\right)  +}\left(  x\right)  \ \ \ \ \ \ \ \ \ \ \text{for every }%
x\in\operatorname*{Ker}\varepsilon.
\]
Then, (\ref{ind2}) shows that%
\[
\varphi_{\infty+}\left(  x\right)  =\varphi_{m+}\left(  x\right)
\ \ \ \ \ \ \ \ \ \ \text{for every }x\in\operatorname*{Ker}\varepsilon\text{
and every }m\in\mathbb{N}\text{ satisfying }x\in\mathcal{H}^{m}.
\]
\newline Now (\ref{ind3}) proves that $\varphi_{\infty+}=\varphi_{\infty-}%
\ast\varphi$.\newline We now extend the map $\varphi_{\infty-}%
:\operatorname*{Ker}\varepsilon\rightarrow\mathcal{A}$ to a map $\varphi
_{-}:\mathcal{H}\rightarrow\mathcal{A}$ by setting $\varphi_{-}\left(
\mathbf{1}\right)  =\mathbf{1}_{\mathcal{A}}$. Similarly, we extend the map
$\varphi_{\infty+}:\operatorname*{Ker}\varepsilon\rightarrow\mathcal{A}$ to a
map $\varphi_{+}:\mathcal{H}\rightarrow\mathcal{A}$ by setting $\varphi
_{+}\left(  \mathbf{1}\right)  =\mathbf{1}_{\mathcal{A}}$. It is easy to see
that $\varphi_{+}=\varphi_{-}\ast\varphi$ (since $\varphi_{\infty+}%
=\varphi_{\infty-}\ast\varphi$ and since $\varphi_{+}\left(  \mathbf{1}%
\right)  =\mathbf{1}_{\mathcal{A}}=\left(  \varphi_{-}\ast\varphi\right)
\left(  \mathbf{1}\right)  $), so that $\varphi=\varphi_{-}^{\ast-1}%
\ast\varphi_{+}$. Also, it is clear that $\varphi_{-}$ sends $\mathbf{1}$ to
$\mathbf{1}_{\mathcal{A}}$ and $\operatorname*{Ker}\varepsilon$ to
$\mathcal{A}_{-}$ (the latter is because of (\ref{rec1}) and because $\pi$ is
a projection onto $\mathcal{A}_{-}$), and that $\varphi_{+}$ sends
$\mathcal{H}$ to $\mathcal{A}_{+}$ (this is because of (\ref{rec2}) and
because $I-\pi$ is a projection onto $\mathcal{A}_{+}$).\newline We have now
proven the existence of the Birkhoff decomposition. To complete the proof of
Theorem II.5.1 part 1), we must now show that it is unique. To show this, we
assume that we have some elements $\psi_{-}$ and $\psi_{+}$ of $G$ satisfying
$\varphi=\psi_{-}^{-1}\ast\psi_{+}$ such that $\psi_{-}$ sends $\mathbf{1}$ to
$\mathbf{1}_{\mathcal{A}}$ and $\operatorname*{Ker}\varepsilon$ to
$\mathcal{A}_{-}$, and such that $\psi_{+}$ sends $\mathcal{H}$ to
$\mathcal{A}_{+}$. Now let us prove that $\psi_{-}=\varphi_{-}$ and $\psi
_{+}=\varphi_{+}$; this will clearly prove the uniqueness of the Birkhoff
decomposition.\newline To prove that $\psi_{-}=\varphi_{-}$, we will show by
induction over $n$ that $\psi_{-}\mid_{\mathcal{H}^{+n}}=\varphi_{n-}$ for
every $n\in\mathbb{N}$: The induction base (the case $n=0$) is clear (again
due to $\mathcal{H}^{+0}=0$). Now to the induction step: Let $m\in\mathbb{N}$
be arbitrary. Assume that $\psi_{-}\mid_{\mathcal{H}^{+m}}=\varphi_{m-}$. Let
us now show that $\psi_{-}\mid_{\mathcal{H}^{+\left(  m+1\right)  }}%
=\varphi_{\left(  m+1\right)  -}$.\newline Let $x\in\mathcal{H}^{+\left(
m+1\right)  }$. Then,%
\[
\sum\limits_{\left(  x\right)  }x^{\prime}\otimes x^{\prime\prime
}=\widetilde{\Delta}x\in\sum\limits_{\substack{p+q=m+1;\\p\neq0;\ q\neq
0}}\underbrace{\mathcal{H}^{+p}}_{\subseteq\mathcal{H}^{+m}}\otimes
\underbrace{\mathcal{H}^{+q}}_{\subseteq\mathcal{H}}\subseteq\sum
\limits_{\substack{p+q=m+1;\\p\neq0;\ q\neq0}}\mathcal{H}^{+m}\otimes
\mathcal{H}\subseteq\mathcal{H}^{+m}\otimes\mathcal{H}.
\]
We can thus WLOG assume that $x^{\prime}\in\mathcal{H}^{+m}$. On the other
hand, $\psi_{-}\left(  x\right)  \in\mathcal{A}_{-}$ (since $x\in
\mathcal{H}^{+\left(  m+1\right)  }\subseteq\operatorname*{Ker}\varepsilon$
and $\psi_{-}\left(  \operatorname*{Ker}\varepsilon\right)  \subseteq
\mathcal{A}_{-}$), so that $\pi\left(  \psi_{-}\left(  x\right)  \right)
=\psi_{-}\left(  x\right)  $ (because $\pi$ is a projection onto
$\mathcal{A}_{-}$). Besides, $\psi_{+}\left(  x\right)  \in\mathcal{A}_{+}$,
and thus $\pi\left(  \psi_{+}\left(  x\right)  \right)  =0$ (since $\pi$ is a
projection parallel to $\mathcal{A}_{+}$). But since $\psi_{+}=\psi_{-}%
\ast\varphi$ (because $\varphi=\psi_{-}^{-1}\ast\psi_{+}$), we have%
\begin{align*}
\psi_{+}\left(  x\right)   &  =\left(  \psi_{-}\ast\varphi\right)  \left(
x\right)  =\sum\limits_{\left(  x\right)  }\psi_{-}\left(  x_{1}\right)
\varphi\left(  x_{2}\right)  =\psi_{-}\left(  x\right)  +\varphi\left(
x\right)  +\sum\limits_{\left(  x\right)  }\underbrace{\psi_{-}\left(
x^{\prime}\right)  }_{\substack{=\left(  \psi_{-}\mid_{\mathcal{H}^{+m}%
}\right)  \left(  x^{\prime}\right)  \\\text{(since }x^{\prime}\in
\mathcal{H}^{+m}\text{)}}}\varphi\left(  x^{\prime\prime}\right) \\
&  =\psi_{-}\left(  x\right)  +\varphi\left(  x\right)  +\sum\limits_{\left(
x\right)  }\underbrace{\left(  \psi_{-}\mid_{\mathcal{H}^{+m}}\right)
}_{=\varphi_{m-}}\left(  x^{\prime}\right)  \varphi\left(  x^{\prime\prime
}\right)  =\psi_{-}\left(  x\right)  +\varphi\left(  x\right)  +\sum
\limits_{\left(  x\right)  }\varphi_{m-}\left(  x^{\prime}\right)
\varphi\left(  x^{\prime\prime}\right)  .
\end{align*}
Thus%
\begin{align*}
\pi\left(  \psi_{+}\left(  x\right)  \right)   &  =\pi\left(  \psi_{-}\left(
x\right)  +\varphi\left(  x\right)  +\sum\limits_{\left(  x\right)  }%
\varphi_{m-}\left(  x^{\prime}\right)  \varphi\left(  x^{\prime\prime}\right)
\right) \\
&  =\underbrace{\pi\left(  \psi_{-}\left(  x\right)  \right)  }_{=\psi
_{-}\left(  x\right)  }+\pi\left(  \varphi\left(  x\right)  +\sum
\limits_{\left(  x\right)  }\varphi_{m-}\left(  x^{\prime}\right)
\varphi\left(  x^{\prime\prime}\right)  \right) \\
&  =\psi_{-}\left(  x\right)  +\pi\left(  \varphi\left(  x\right)
+\sum\limits_{\left(  x\right)  }\varphi_{m-}\left(  x^{\prime}\right)
\varphi\left(  x^{\prime\prime}\right)  \right)  .
\end{align*}
Since $\pi\left(  \psi_{+}\left(  x\right)  \right)  =0$, this becomes
$0=\psi_{-}\left(  x\right)  +\pi\left(  \varphi\left(  x\right)
+\sum\limits_{\left(  x\right)  }\varphi_{m-}\left(  x^{\prime}\right)
\varphi\left(  x^{\prime\prime}\right)  \right)  $, so that%
\[
\psi_{-}\left(  x\right)  =-\pi\left(  \varphi\left(  x\right)  +\sum
\limits_{\left(  x\right)  }\varphi_{m-}\left(  x^{\prime}\right)
\varphi\left(  x^{\prime\prime}\right)  \right)  =\varphi_{\left(  m+1\right)
-}\left(  x\right)  .
\]
Since this holds for every $x\in\mathcal{H}^{+\left(  m+1\right)  }$, we thus
conclude that $\psi_{-}\mid_{\mathcal{H}^{+\left(  m+1\right)  }}%
=\varphi_{\left(  m+1\right)  -}$. This completes the induction.\newline We
thus have shown that $\psi_{-}\mid_{\mathcal{H}^{+n}}=\varphi_{n-}$ for every
$n\in\mathbb{N}$. By the construction of $\varphi_{\infty-}$, this yields that
$\psi_{-}\mid_{\operatorname*{Ker}\varepsilon}=\varphi_{\ast-}$. This means
that the maps $\psi_{-}$ and $\varphi_{-}$ coincide on $\operatorname*{Ker}%
\varepsilon$. Since they also coincide on $k\cdot\mathbf{1}$, this yields
$\psi_{-}=\varphi_{-}$. Analogous arguments show that $\psi_{+}=\varphi_{+}$.
This completes the proof of the uniqueness of the Birkhoff decomposition.
Thus, part 1) of Theorem II.5.1 is finally proven.

\item \textbf{Proof of Theorem II.5.1:} On page 36, you write: "The same
property for $\tau_{+}$ comes then from proposition II.3.1." Maybe you mean
Proposition II.4.1 instead of II.3.1 here?

\item \textbf{Proof of Theorem II.5.1:} On page 36, you prove $\tau_{-}\left(
xy\right)  =\tau_{-}\left(  yx\right)  $ by decomposing $\Delta\left(
xy\right)  $ using Sweedler's notation. This decomposition is only correct
when $x$ and $y$ lie in $\operatorname*{Ker}\varepsilon$. (Fortunately, it is
enough to prove $\tau_{-}\left(  xy\right)  =\tau_{-}\left(  yx\right)  $ for
$x$ and $y$ lying in $\operatorname*{Ker}\varepsilon$, because it is trivially
true when $x$ or $y$ lies in $k\cdot\mathbf{1}$.)

\item \textbf{Proof of Theorem II.5.1:} On page 37, you write "with
$X=\chi\left(  x\right)  -\sum_{\left(  x\right)  }\chi_{-}\left(  x^{\prime
}\right)  \chi\left(  x^{\prime\prime}\right)  $ and $Y=\chi\left(  y\right)
-\sum_{\left(  y\right)  }\chi_{-}\left(  y^{\prime}\right)  \chi\left(
y^{\prime\prime}\right)  $" (during the proof of assertion 3)). This should be
"with $X=\chi\left(  x\right)  +\sum_{\left(  x\right)  }\chi_{-}\left(
x^{\prime}\right)  \chi\left(  x^{\prime\prime}\right)  $ and $Y=\chi\left(
y\right)  +\sum_{\left(  y\right)  }\chi_{-}\left(  y^{\prime}\right)
\chi\left(  y^{\prime\prime}\right)  $".

\item \textbf{Remark at the end of II.5:} Remove the word "recursively": the
current definition of $b$ is not recursive at all. Besides, $\varphi_{-}%
=-\pi\circ b\left(  \varphi\right)  $ should probably be replaced by
$\varphi_{-}=e-\pi\circ b\left(  \varphi\right)  $, unless you want it to hold
on $\operatorname*{Ker}\varepsilon$ only (since $\varphi_{-}\left(
\mathbf{1}\right)  =\mathbf{1}_{\mathcal{A}}\notin\pi\left(  \mathcal{A}%
\right)  $).

\item \textbf{II.6:} At the very end of page 39, you write: "K. Ebrahimi-Fard,
L. Guo and D. Kreimer derive in [EGK2] two identities involving the Bogoliubov
character :%
\[
e^{\ast-R\left(  \chi_{R}\left(  X\right)  \right)  }=R\left(  b\left(
e^{\ast X}\right)  \right)  ,\ \ \ \ \ \ \ \ \ \ e^{\ast\widetilde{R}\left(
\chi_{R}\left(  X\right)  \right)  }=-\widetilde{R}\left(  b\left(  e^{\ast
X}\right)  \right)  .
\]
" I have not read [EGK2], but I think that by applying the uniqueness of the
Birkhoff decomposition and the (corrected!) version of the Remark at the end
of II.5 ("corrected" in the sense that $\varphi_{-}=-\pi\circ b\left(
\varphi\right)  $ is replaced by $\varphi_{-}=e-\pi\circ b\left(
\varphi\right)  $), I get two slightly different identities:%
\[
e^{\ast-R\left(  \chi_{R}\left(  X\right)  \right)  }=e-R\left(  b\left(
e^{\ast X}\right)  \right)  ,\ \ \ \ \ \ \ \ \ \ e^{\ast\widetilde{R}\left(
\chi_{R}\left(  X\right)  \right)  }=\widetilde{R}\left(  b\left(  e^{\ast
X}\right)  \right)  .
\]


\item \textbf{Remark at the end of II.6:} What do you mean by "Rota-Baxter
identity for $R$ just guarantees that equation (**) gives a Birkhoff
decomposition"? How do you define a Birkhoff decomposition when there is no
projection $\pi$ but just an operator $R$ satisfying Rota-Baxter?

\item \textbf{II.7:} Replace "fuction" by "function".

\item \textbf{II.8:} On the first line of II.8, you speak of "biderivation".
What does "biderivation" mean? I used to understand this word as "derivation
and coderivation at the same time", but the meaning of "derivation" that makes
$Y$ a derivation (and makes $\varphi\mapsto\varphi\circ Y$ a derivation of
$\left(  \mathcal{L}\left(  \mathcal{H},\mathcal{A}\right)  ,\ast\right)  $ in
Lemma II.8.1) is different from the meaning of "derivation" in II.4, so you
might want to add a remark about these two meanings (the first one defines a
derivation as a map $f$ satisfying $f\left(  xy\right)  =xf\left(  y\right)
+f\left(  x\right)  y$, whereas the second one defines a derivation as a map
$f$ satisfying $f\left(  xy\right)  =e\left(  x\right)  f\left(  y\right)
+f\left(  x\right)  e\left(  y\right)  $) and which of them is used at what
place. Also, you have never defined what a coderivation is.

\item \textbf{II.8:} Before Lemma II.8.1, you define a map $\theta_{t}$. Maybe
you should add that $t$ is assumed to be an element for which $e^{nt}$ makes
sense, e. g., a complex number, or an element of a maximal ideal in a complete
local ring.

\item \textbf{Lemma II.8.2:} This lemma is not completely proved in the text
preceding it. What remains to be proven is that if $\alpha$ is a derivation of
$\mathcal{H}$ with values in $\mathcal{A}$, then $\alpha\circ Y^{-1}$ is a
derivation of $\mathcal{H}$ with values in $\mathcal{A}$ as well.\newline Here
is a \textit{proof} for this: Assume that $\alpha$ is a derivation of
$\mathcal{H}$ with values in $\mathcal{A}$. Now, in order to prove that
$\alpha\circ Y^{-1}$ is a derivation of $\mathcal{H}$ with values in
$\mathcal{A}$, we let $x$ and $y$ be two elements of $\mathcal{H}$. We must
then prove the equation $\left(  \alpha\circ Y^{-1}\right)  \left(  xy\right)
=e\left(  x\right)  \left(  \alpha\circ Y^{-1}\right)  \left(  y\right)
+\left(  \alpha\circ Y^{-1}\right)  \left(  x\right)  e\left(  y\right)
$.\newline Since this equation is linear in $x$ and $y$, we can WLOG assume
that $x$ and $y$ are homogeneous. We now distinguish between four
cases:\newline\textit{Case 1:} We have $\left\vert x\right\vert >0$ and
$\left\vert y\right\vert >0$.\newline\textit{Case 2:} We have $\left\vert
x\right\vert =0$ and $\left\vert y\right\vert >0$.\newline\textit{Case 3:} We
have $\left\vert y\right\vert =0$ and $\left\vert x\right\vert >0$%
.\newline\textit{Case 4:} We have $\left\vert x\right\vert =0$ and $\left\vert
y\right\vert =0$.\newline Let us consider Case 1 first. In this case,
\begin{align*}
\left(  \alpha\circ Y^{-1}\right)  \left(  xy\right)   &  =\alpha\left(
\underbrace{Y^{-1}\left(  xy\right)  }_{=\dfrac{1}{\left\vert xy\right\vert
}xy}\right)  =\dfrac{1}{\left\vert xy\right\vert }\underbrace{\alpha\left(
xy\right)  }_{\substack{=e\left(  x\right)  \alpha\left(  y\right)
+\alpha\left(  x\right)  e\left(  y\right)  \\\text{(since }\alpha\text{ is a
derivation)}}}\\
&  =\dfrac{1}{\left\vert xy\right\vert }\left(  \underbrace{e\left(  x\right)
}_{=0\text{ (since }\left\vert x\right\vert >0\text{)}}\alpha\left(  y\right)
+\alpha\left(  x\right)  \underbrace{e\left(  y\right)  }_{=0\text{ (since
}\left\vert y\right\vert >0\text{)}}\right) \\
&  =\dfrac{1}{\left\vert xy\right\vert }\left(  0\alpha\left(  y\right)
+\alpha\left(  x\right)  0\right)  =0
\end{align*}
and%
\[
\underbrace{e\left(  x\right)  }_{=0\text{ (since }\left\vert x\right\vert
>0\text{)}}\left(  \alpha\circ Y^{-1}\right)  \left(  y\right)  +\left(
\alpha\circ Y^{-1}\right)  \left(  x\right)  \underbrace{e\left(  y\right)
}_{=0\text{ (since }\left\vert y\right\vert >0\text{)}}=0\left(  \alpha\circ
Y^{-1}\right)  \left(  y\right)  +\left(  \alpha\circ Y^{-1}\right)  \left(
x\right)  0=0,
\]
so that $\left(  \alpha\circ Y^{-1}\right)  \left(  xy\right)  =e\left(
x\right)  \left(  \alpha\circ Y^{-1}\right)  \left(  y\right)  +\left(
\alpha\circ Y^{-1}\right)  \left(  x\right)  e\left(  y\right)  $ is proven in
Case 1.\newline Let us now consider Case 2. In this case,%
\begin{align*}
\left(  \alpha\circ Y^{-1}\right)  \left(  xy\right)   &  =\alpha\left(
\underbrace{Y^{-1}\left(  xy\right)  }_{=\dfrac{1}{\left\vert xy\right\vert
}xy}\right)  =\dfrac{1}{\left\vert xy\right\vert }\underbrace{\alpha\left(
xy\right)  }_{\substack{=e\left(  x\right)  \alpha\left(  y\right)
+\alpha\left(  x\right)  e\left(  y\right)  \\\text{(since }\alpha\text{ is a
derivation)}}}=\dfrac{1}{\left\vert xy\right\vert }\left(  e\left(  x\right)
\alpha\left(  y\right)  +\alpha\left(  x\right)  \underbrace{e\left(
y\right)  }_{=0\text{ (since }\left\vert y\right\vert >0\text{)}}\right) \\
&  =\dfrac{1}{\left\vert xy\right\vert }\left(  e\left(  x\right)
\alpha\left(  y\right)  +\alpha\left(  x\right)  0\right)  =\dfrac
{1}{\left\vert xy\right\vert }e\left(  x\right)  \alpha\left(  y\right)
=\dfrac{1}{\left\vert y\right\vert }e\left(  x\right)  \alpha\left(  y\right)
\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\left\vert xy\right\vert
=\underbrace{\left\vert x\right\vert }_{=0}+\left\vert y\right\vert
=\left\vert y\right\vert \right)
\end{align*}
and%
\begin{align*}
&  e\left(  x\right)  \left(  \alpha\circ Y^{-1}\right)  \left(  y\right)
+\left(  \alpha\circ Y^{-1}\right)  \left(  x\right)  \underbrace{e\left(
y\right)  }_{=0\text{ (since }\left\vert y\right\vert >0\text{)}}\\
&  =e\left(  x\right)  \left(  \alpha\circ Y^{-1}\right)  \left(  y\right)
+\left(  \alpha\circ Y^{-1}\right)  \left(  x\right)  0=e\left(  x\right)
\underbrace{\left(  \alpha\circ Y^{-1}\right)  \left(  y\right)  }%
_{=\alpha\left(  Y^{-1}\left(  y\right)  \right)  =\alpha\left(  \dfrac
{1}{\left\vert y\right\vert }y\right)  =\dfrac{1}{\left\vert y\right\vert
}\alpha\left(  y\right)  }=\dfrac{1}{\left\vert y\right\vert }e\left(
x\right)  \alpha\left(  y\right)  ,
\end{align*}
so that $\left(  \alpha\circ Y^{-1}\right)  \left(  xy\right)  =e\left(
x\right)  \left(  \alpha\circ Y^{-1}\right)  \left(  y\right)  +\left(
\alpha\circ Y^{-1}\right)  \left(  x\right)  e\left(  y\right)  $ is proven in
Case 2.\newline Similarly we can prove $\left(  \alpha\circ Y^{-1}\right)
\left(  xy\right)  =e\left(  x\right)  \left(  \alpha\circ Y^{-1}\right)
\left(  y\right)  +\left(  \alpha\circ Y^{-1}\right)  \left(  x\right)
e\left(  y\right)  $ in Case 3.\newline In Case 4, the proof is left to the
reader.\newline Thus, in all four cases, we have shown that $\left(
\alpha\circ Y^{-1}\right)  \left(  xy\right)  =e\left(  x\right)  \left(
\alpha\circ Y^{-1}\right)  \left(  y\right)  +\left(  \alpha\circ
Y^{-1}\right)  \left(  x\right)  e\left(  y\right)  $, so that we can conclude
that $\alpha\circ Y^{-1}$ is a derivation, qed.

\item \textbf{Remark at the end of II.8:} The comment that "any other value of
$Y^{-1}\left(  \mathbf{1}\right)  $ would give the same result" is correct
only as long as this value of $Y^{-1}\left(  \mathbf{1}\right)  $ is still
supposed to be a scalar multiple of $\mathbf{1}$.

\item \textbf{II.9.1:} You write: "and we extend $\Delta$ to an algebra
isomorphism". You mean "homomorphism", not "isomorphism".

\item \textbf{II.9.1:} It would be useful to add that this Hopf algebra
$\mathcal{N}$ is isomorphic to the symmetric Hopf algebra on the $k$-module
$k^{\mathbb{P}}$ (where $\mathbb{P}$ is the set of all primes) by the
isomorphism%
\begin{align*}
\mathcal{N}  &  \rightarrow k^{\mathbb{P}};\\
e_{n}  &  \mapsto\sum_{p\text{ prime divisor of }n}\left(  \text{multiplicity
of }p\text{ in }n\right)  \cdot\left(  \text{basis vector of }k^{\mathbb{P}%
}\text{ corresponding to }p\right)  .
\end{align*}


\item \textbf{II.9.3:} The definition you give for the notion of a "planar
rooted tree" confused me. If it is taken literally, two planar graphs are
considered different if they differ in the position of their vertices on the
plane; for instance, two planar rooted trees consisting of one vertex each are
different if these vertices are located at different points in the plane. I
\textit{believe} this is not what you want to achieve. After having looked up
some definitions of "planar rooted tree" on the internet, I would say I prefer
the purely combinatorial definition: A planar rooted tree is an oriented tree
(this means an oriented connected graph with no oriented cycles such that only
one vertex - the root - has only outgoing edges, whereas every other vertex
has exactly one incoming edge) along with, for every vertex $v$ of the tree, a
linear order on the set of children of $v$ (here, a "child" means the target
of an outgoing edge from $v$). When we draw a planar rooted tree on plane, we
have to start with the root, and at each step choose a vertex already drawn
and add all of its children; we should do it in such a way that the lowest
child (with respect to the linear order) is drawn on the very left, the second
lowest child is the second from the left, and so on, and the highest child is
drawn on the very right.\newline Also, when you say "Let $\mathcal{T}$ be the
set of planar rooted trees", you mean "Let $\mathcal{T}$ be the set of planar
rooted trees up to isomorphism".\newline Also, it should be mentioned that a
"planar rooted forest" means a finite (possibly empty) sequence of planar
rooted trees. This means that changing the order of the trees in a planar
rooted forest changes the forest.

\item \textbf{II.9.3:} You write: "The trunk of a tree is a tree". This is
only true if the cut is not the empty cut (since there is no tree with $0$ vertices).

\item \textbf{II.9.3:} You write: "and let $\operatorname*{Adm}\nolimits^{\ast
}\left(  F\right)  $ the set of elementary cuts disregarding the empty cut and
the total cut". You mean "admissible" rather than "elementary" here.

\item \textbf{II.9.3:} I would prefer to have a formal definition of a notion
of a "cut", particularly because there is a very simple one:\newline- A
\textit{cut} of a forest $F$ denotes a pair $\left(  A,B\right)  $ where $A$
and $B$ are two subsets of $\operatorname*{V}\left(  F\right)  $ (the set of
the vertices of $F$) such that $A\cap B=\varnothing$ and $A\cup
B=\operatorname*{V}\left(  F\right)  $. The \textit{trunk} of $F$ with respect
to this cut is then defined as the forest $\operatorname*{ind}\nolimits_{A}%
\left(  F\right)  $, and the \textit{crown} of $F$ with respect to this cut is
defined as the forest $\operatorname*{ind}\nolimits_{B}\left(  F\right)
$.\newline Here, whenever $S$ is a subset of $\operatorname*{V}\left(
F\right)  $, we denote by $\operatorname*{ind}\nolimits_{S}\left(  F\right)  $
the induced subgraph of $F$ on the vertex set $S$.\newline With this
definition, we can reformulate the definitions of admissible and elementary
cuts, as well as of bi-admissible couples, in a more formal fashion:\newline-
An \textit{admissible cut} of a forest $F$ denotes a cut $\left(  A,B\right)
$ such that every path on $F$ (viewed as a sequence of vertices) has the form
$\left(  \text{some vertices in }A\text{,\ some vertices in }B\right)  $
(where "some" might also mean "none"). Equivalently, an \textit{admissible
cut} of a forest $F$ denotes a cut $\left(  A,B\right)  $ such that the set
$A$ is closed under taking ancestors (i. e., every ancestor of an element of
$A$ lies in $A$) and the set $B$ is closed under taking descendants (i. e.,
every descendant of an element of $B$ lies in $B$).\newline- An
\textit{elementary cut} of a forest $F$ denotes a cut $\left(  A,B\right)  $
such that $B=\left(  \text{the set of all descendants of }v\right)  $ for some
$v\in\operatorname*{V}\left(  F\right)  $. Here, $v$ is considered its own
descendant.\newline- A \textit{bi-admissible couple} means a triple $\left(
A_{1},A_{2},A_{3}\right)  $ where $A_{1}$, $A_{2}$ and $A_{3}$ are three
subsets of $\operatorname*{V}\left(  F\right)  $ (the set of the vertices of
$F$) such that%
\[
A_{i}\cap A_{j}=\varnothing\text{ for all }i,j\in\left\{  1,2,3\right\}
\text{ with }i\neq j
\]
and $A_{1}\cup A_{2}\cup A_{3}=\operatorname*{V}\left(  F\right)  $, and such
that every $i,j\in\left\{  1,2,3\right\}  $, every vertex $v\in A_{i}$ and
every descendant $w\in A_{j}$ of $v$ satisfy $i\leq j$. The \textit{trunk} of
$F$ with respect to this bi-admissible couple is defined as the forest
$\operatorname*{ind}\nolimits_{A}\left(  F\right)  $, the \textit{middle} of
$F$ with respect to this bi-admissible couple is defined as the forest
$\operatorname*{ind}\nolimits_{B}\left(  F\right)  $, and the \textit{crown}
of $F$ with respect to this bi-admissible couple is defined as the forest
$\operatorname*{ind}\nolimits_{C}\left(  F\right)  $.\newline It is then easy
to see that choosing a bi-admissible couple on a forest $F$ is equivalent to
choosing an admissible cut on $F$ and then choosing an admissible cut on the
crown of $F$ (indeed, a triple $\left(  A_{1},A_{2},A_{3}\right)  $ of subsets
of $\operatorname*{V}\left(  F\right)  $ is a bi-admissible couple if and only
if $\left(  A_{1},A_{2}\cup A_{3}\right)  $ is an admissible cut of $F$ and
$\left(  A_{2},A_{3}\right)  $ is an admissible cut of $\operatorname*{ind}%
\nolimits_{A_{2}\cup A_{3}}\left(  F\right)  $), and also equivalent to
choosing an admissible cut on $F$ and then choosing an admissible cut on the
trunk of $F$ (indeed, a triple $\left(  A_{1},A_{2},A_{3}\right)  $ of subsets
of $\operatorname*{V}\left(  F\right)  $ is a bi-admissible couple if and only
if $\left(  A_{1}\cup A_{2},A_{3}\right)  $ is an admissible cut of $F$ and
$\left(  A_{1},A_{2}\right)  $ is an admissible cut of $\operatorname*{ind}%
\nolimits_{A_{1}\cup A_{2}}\left(  F\right)  $). This proves the
coassociativity of $\Delta$.

\item \textbf{II.9.3:} On page 44, replace "n-uples" by "$n$-uples" (the "n"
should be math, not text).

\item \textbf{II.9.3:} On page 44, "By corollary II.2.2" should be "By
corollary II.3.2".

\item \textbf{III.3:} On the fourth line of page 47 (not counting the title of
III.3), you write: "the free commutative algebra generated by $V$". Is it
possible that you mean $V_{\mathcal{T}}$ when you say $V$ ?

\item \textbf{III.3:} On the sixth line of page 47, replace "co-unity" by "co-unit".

\item \textbf{III.3:} On page 47, you write that "any nonempty subgraph has a
non-vanishing loop number". How can this be true? For example, assume that our
original graph is a $4$-cycle, and we choose the subgraph formed by two
adjacent edges of this cycle. This subgraph has $2$ internal edges, $2$
external edges and $3$ vertices, so the loop number is $2-3+1=0$. Maybe you
want the $I$ in the definition of the loop number to be the number of
\textit{all} edges?

\item \textbf{III.3:} In the fourth line from bottom of page 48, you write:
"The ideal $J$ is the a bi-ideal". Here, the "the" should be removed.

\item \textbf{III.3:} On page 49, you write: "We can identify the quotient
with $S\left(  V_{T}^{\prime}\right)  $, where $V_{T}^{\prime}$ stands for
[...]". Both of the $T$'s here should be calligraphic $\mathcal{T}$'s.

\item \textbf{Chapter IV.} On page 50, you write: "We denote by $Y$ (resp.
$\theta_{t}$) the biderivation (resp. the one-parameter group of
automorphisms) of the Hopf algebra $\mathcal{H}$ induced by the graduation
(cf. \S \ II.6)." I think you mean II.8, not II.6 here.

\item \textbf{Proof of Proposition IV.1.2:} You mention a "derivation property%
\[
\left(  e^{\ast t\alpha}\ast e^{\ast s\alpha}\right)  \circ Y=\left(  e^{\ast
t\alpha}\circ Y\right)  \ast e^{\ast s\alpha}+e^{\ast t\alpha}\ast\left(
e^{\ast s\alpha}\circ Y\right)
\]
" here. It is correct, but it will probably become a bit clearer if you refer
to Lemma II.8.1 for this property (otherwise, it looks like you are applying
the fact that $Y$ is a derivation).

\item \textbf{Proof of Corollary IV.1.3:} I don't understand how exactly you
show the first assertion here.\newline Anyway, I don't like the proof as it
uses the analytic Proposition IV.1.2, whereas there is a much more
straightforward proof using pure algebra (and therefore works for any
commutative $k$-algebra $\mathcal{A}$ over any field $k$, not necessarily
$\mathbb{C}$):\newline\textit{Alternative proof of Corollary IV.1.3:} Let us
prove the first assertion first:\newline Let $\alpha$ be a derivation of
$\mathcal{H}$ with values in $\mathcal{A}$. We must prove that $R\left(
\alpha\right)  $ is a derivation of $\mathcal{H}$ with values in $\mathcal{A}$
as well.\newline Let $\varphi=e^{\ast\alpha}$. Then, Proposition II.4.2 yields
that $\varphi\in G_{1}$ (since $\alpha$ is a derivation and thus $\alpha
\in\mathfrak{g}_{1}$), so that $\varphi$ is an algebra homomorphism. Thus,
$\varphi^{\ast-1}$ is an algebra homomorphism as well (since $G_{1}$ is a
group). Now, we are going to prove that any $x\in\mathcal{H}$ and
$y\in\mathcal{H}$ satisfy%
\begin{equation}
\left(  \varphi^{\ast-1}\ast\left(  \varphi\circ Y\right)  \right)  \left(
xy\right)  =e\left(  x\right)  \left(  \varphi^{\ast-1}\ast\left(
\varphi\circ Y\right)  \right)  \left(  y\right)  +\left(  \varphi^{\ast
-1}\ast\left(  \varphi\circ Y\right)  \right)  \left(  x\right)  e\left(
y\right)  . \label{cor1}%
\end{equation}
\newline In order to prove (\ref{cor1}), we can WLOG assume that $x$ and $y$
are homogeneous. Then, we can write $\Delta\left(  x\right)  =\sum
\limits_{\left(  x\right)  }x_{1}\otimes x_{2}$ with homogeneous $x_{1}$ and
$x_{2}$, and we can write $\Delta\left(  y\right)  =\sum\limits_{\left(
y\right)  }y_{1}\otimes y_{2}$ with homogeneous $y_{1}$ and $y_{2}$. Clearly,%
\begin{align*}
\left(  \varphi^{\ast-1}\ast\left(  \varphi\circ Y\right)  \right)  \left(
x\right)   &  =\sum\limits_{\left(  x\right)  }\varphi^{\ast-1}\left(
x_{1}\right)  \underbrace{\left(  \varphi\circ Y\right)  \left(  x_{2}\right)
}_{\substack{=\varphi\left(  Y\left(  x_{2}\right)  \right)  =\varphi\left(
\left\vert x_{2}\right\vert x_{2}\right)  \\\text{(since }Y\left(
x_{2}\right)  =\left\vert x_{2}\right\vert x_{2}\text{)}}}=\sum
\limits_{\left(  x\right)  }\varphi^{\ast-1}\left(  x_{1}\right)
\varphi\left(  \left\vert x_{2}\right\vert x_{2}\right) \\
&  =\sum\limits_{\left(  x\right)  }\left\vert x_{2}\right\vert \varphi
^{\ast-1}\left(  x_{1}\right)  \varphi\left(  x_{2}\right)  .
\end{align*}
Similarly, $\left(  \varphi^{\ast-1}\ast\left(  \varphi\circ Y\right)
\right)  \left(  y\right)  =\sum\limits_{\left(  y\right)  }\left\vert
y_{2}\right\vert \varphi^{\ast-1}\left(  y_{1}\right)  \varphi\left(
y_{2}\right)  $. But%
\begin{align*}
&  \left(  \varphi^{\ast-1}\ast\left(  \varphi\circ Y\right)  \right)  \left(
xy\right) \\
&  =\sum_{\left(  xy\right)  }\varphi^{\ast-1}\left(  \left(  xy\right)
_{1}\right)  \left(  \varphi\circ Y\right)  \left(  \left(  xy\right)
_{2}\right)  =\sum\limits_{\left(  x\right)  \left(  y\right)  }%
\underbrace{\varphi^{\ast-1}\left(  x_{1}y_{1}\right)  }_{\substack{=\varphi
^{\ast-1}\left(  x_{1}\right)  \varphi^{\ast-1}\left(  y_{1}\right)
\\\text{(since }\varphi^{\ast-1}\text{ is an algebra}\\\text{homomorphism)}%
}}\underbrace{\left(  \varphi\circ Y\right)  \left(  x_{2}y_{2}\right)
}_{=\varphi\left(  Y\left(  x_{2}y_{2}\right)  \right)  }\\
&  =\sum\limits_{\left(  x\right)  \left(  y\right)  }\varphi^{\ast-1}\left(
x_{1}\right)  \varphi^{\ast-1}\left(  y_{1}\right)  \varphi\left(
\underbrace{Y\left(  x_{2}y_{2}\right)  }_{=\left(  \left\vert x_{2}%
\right\vert +\left\vert y_{2}\right\vert \right)  x_{2}y_{2}}\right)
=\sum\limits_{\left(  x\right)  \left(  y\right)  }\left(  \left\vert
x_{2}\right\vert +\left\vert y_{2}\right\vert \right)  \varphi^{\ast-1}\left(
x_{1}\right)  \varphi^{\ast-1}\left(  y_{1}\right)  \underbrace{\varphi\left(
x_{2}y_{2}\right)  }_{\substack{=\varphi\left(  x_{2}\right)  \varphi\left(
y_{2}\right)  \\\text{(since }\varphi\text{ is an algebra}%
\\\text{homomorphism)}}}\\
&  =\sum\limits_{\left(  x\right)  \left(  y\right)  }\underbrace{\left(
\left\vert x_{2}\right\vert +\left\vert y_{2}\right\vert \right)
\varphi^{\ast-1}\left(  x_{1}\right)  \varphi^{\ast-1}\left(  y_{1}\right)
\varphi\left(  x_{2}\right)  \varphi\left(  y_{2}\right)  }_{=\left\vert
x_{2}\right\vert \varphi^{\ast-1}\left(  x_{1}\right)  \varphi\left(
x_{2}\right)  \varphi^{\ast-1}\left(  y_{1}\right)  \varphi\left(
y_{2}\right)  +\left\vert y_{2}\right\vert \varphi^{\ast-1}\left(
x_{1}\right)  \varphi\left(  x_{2}\right)  \varphi^{\ast-1}\left(
y_{1}\right)  \varphi\left(  y_{2}\right)  }\\
&  =\sum\limits_{\left(  x\right)  \left(  y\right)  }\left(  \left\vert
x_{2}\right\vert \varphi^{\ast-1}\left(  x_{1}\right)  \varphi\left(
x_{2}\right)  \varphi^{\ast-1}\left(  y_{1}\right)  \varphi\left(
y_{2}\right)  +\left\vert y_{2}\right\vert \varphi^{\ast-1}\left(
x_{1}\right)  \varphi\left(  x_{2}\right)  \varphi^{\ast-1}\left(
y_{1}\right)  \varphi\left(  y_{2}\right)  \right) \\
&  =\underbrace{\sum_{\left(  x\right)  }\left\vert x_{2}\right\vert
\varphi^{\ast-1}\left(  x_{1}\right)  \varphi\left(  x_{2}\right)  }_{=\left(
\varphi^{\ast-1}\ast\left(  \varphi\circ Y\right)  \right)  \left(  x\right)
}\underbrace{\sum_{\left(  y\right)  }\varphi^{\ast-1}\left(  y_{1}\right)
\varphi\left(  y_{2}\right)  }_{=\left(  \varphi^{\ast-1}\ast\varphi\right)
\left(  y\right)  =e\left(  y\right)  }+\underbrace{\sum_{\left(  x\right)
}\varphi^{\ast-1}\left(  x_{1}\right)  \varphi\left(  x_{2}\right)
}_{=\left(  \varphi^{\ast-1}\ast\varphi\right)  \left(  x\right)  =e\left(
x\right)  }\underbrace{\sum_{\left(  y\right)  }\left\vert y_{2}\right\vert
\varphi^{\ast-1}\left(  y_{1}\right)  \varphi\left(  y_{2}\right)  }_{=\left(
\varphi^{\ast-1}\ast\left(  \varphi\circ Y\right)  \right)  \left(  y\right)
}\\
&  =\left(  \varphi^{\ast-1}\ast\left(  \varphi\circ Y\right)  \right)
\left(  x\right)  e\left(  y\right)  +e\left(  x\right)  \left(  \varphi
^{\ast-1}\ast\left(  \varphi\circ Y\right)  \right)  \left(  y\right)  .
\end{align*}
This proves (\ref{cor1}). Thus, $\varphi^{\ast-1}\ast\left(  \varphi\circ
Y\right)  $ is a derivation from $\mathcal{H}$ with values in $\mathcal{A}$.
But since $\varphi=e^{\ast\alpha}$ yields $\varphi^{\ast-1}\ast\left(
\varphi\circ Y\right)  =e^{\ast-\alpha}\ast\left(  e^{\ast\alpha}\circ
Y\right)  =R\left(  \alpha\right)  $, this means that $R\left(  \alpha\right)
$ is a derivation from $\mathcal{H}$ with values in $\mathcal{A}$. We thus
have shown the first assertion of Corollary IV.1.3.\newline Let us now prove
the second assertion of Corollary IV.1.3:\newline Let $\beta$ be a cocycle
from $\mathcal{H}$ to $\mathcal{A}$. We must prove that $R\left(
\beta\right)  $ is a cocycle from $\mathcal{H}$ to $\mathcal{A}$ as
well.\newline For this we need a \textit{lemma:} If $\eta$ is a cocycle from
$\mathcal{H}$ to $\mathcal{A}$, then $\eta\circ Y$ is a cocycle from
$\mathcal{H}$ to $\mathcal{A}$ as well.\newline\textit{Proof of the lemma:}
Let $x$ and $y$ be two homogeneous elements of $\mathcal{H}$. Then, $Y\left(
xy\right)  =\underbrace{\left\vert xy\right\vert }_{=\left\vert x\right\vert
+\left\vert y\right\vert }xy=\left(  \left\vert x\right\vert +\left\vert
y\right\vert \right)  xy$ and
\[
\left(  \eta\circ Y\right)  \left(  xy\right)  =\eta\left(
\underbrace{Y\left(  xy\right)  }_{=\left(  \left\vert x\right\vert
+\left\vert y\right\vert \right)  xy}\right)  =\eta\left(  \left(  \left\vert
x\right\vert +\left\vert y\right\vert \right)  xy\right)  =\left(  \left\vert
x\right\vert +\left\vert y\right\vert \right)  \eta\left(  xy\right)  .
\]
Similarly, $\left(  \eta\circ Y\right)  \left(  yx\right)  =\left(  \left\vert
y\right\vert +\left\vert x\right\vert \right)  \eta\left(  yx\right)  $.
Comparing these two equalities (and recalling that $\eta\left(  xy\right)
=\eta\left(  yx\right)  $ (since $\eta$ is a cocyle) and $\left\vert
x\right\vert +\left\vert y\right\vert =\left\vert y\right\vert +\left\vert
x\right\vert $), we conclude that $\left(  \eta\circ Y\right)  \left(
xy\right)  =\left(  \eta\circ Y\right)  \left(  yx\right)  $. We have thus
shown the identity $\left(  \eta\circ Y\right)  \left(  xy\right)  =\left(
\eta\circ Y\right)  \left(  yx\right)  $ for any two homogeneous elements $x$
and $y$ of $\mathcal{H}$. Since this identity is linear in $x$ and $y$, we can
thus conclude that this identity holds for \textit{any two} (not necessarily
homogeneous) elements $x$ and $y$ of $\mathcal{H}$. In other words, $\eta\circ
Y$ is a cocyle. The lemma is proven.\newline Now, Proposition II.4.2 yields
$\exp\left(  \mathfrak{g}_{2}\right)  =G_{2}$. Thus, $\beta\in\mathfrak{g}%
_{2}$ (since $\beta$ is a cocycle) yields $e^{\ast\beta}\in G_{2}$. Since
$G_{2}$ is a group, this yields $e^{\ast-\beta}\in G_{2}$. On the other hand,
the lemma (applied to $Y=e^{\ast\beta}$) yields that $e^{\ast\beta}\circ Y$ is
a cocycle. Now, Proposition II.4.1 1) shows that the convolution
$e^{\ast-\beta}\ast\left(  e^{\ast\beta}\circ Y\right)  $ is a cocycle. Since
$R\left(  \beta\right)  =e^{\ast-\beta}\ast\left(  e^{\ast\beta}\circ
Y\right)  $, this means that $R\left(  \beta\right)  $ is a cocycle. We haved
thus proven the second assertion of Corollary IV.1.3, all without using analysis.

\item \textbf{Remark at the end of IV.1:} Apparently you use Proposition
IV.1.2 to show that $R\left(  \alpha\right)  =\alpha\circ Y$ in the case when
$\mathcal{H}$ is cocommutative. Here is an alternative proof of this
statement:\newline\textit{Alternative proof of }$R\left(  \alpha\right)
=\alpha\circ Y$ \textit{in the case of a cocommutative Hopf algebra
}$\mathcal{H}$\textit{:}\newline From Lemma II.8.1 we know that $\varphi
\mapsto\varphi\circ Y$ is a derivation of the algebra $\left(  \mathcal{L}%
\left(  \mathcal{H},\mathcal{A}\right)  ,\ast\right)  $. In other words, any
$\varphi\in\mathcal{L}\left(  \mathcal{H},\mathcal{A}\right)  $ and $\psi
\in\mathcal{L}\left(  \mathcal{H},\mathcal{A}\right)  $ satisfy $\left(
\varphi\ast\psi\right)  \circ Y=\varphi\ast\left(  \psi\circ Y\right)
+\left(  \varphi\circ Y\right)  \ast\psi$. Applying this property and using
induction over $n$, we can show that every $\alpha\in\mathcal{L}\left(
\mathcal{H},\mathcal{A}\right)  $ and every integer $n>0$ satisfy
$\alpha^{\ast n}\circ Y=n\left(  \alpha\circ Y\right)  \ast\alpha^{\ast\left(
n-1\right)  }$ (here we use that $\mathcal{L}\left(  \mathcal{H}%
,\mathcal{A}\right)  $ is commutative, which follows from the cocommutativity
of $\mathcal{H}$ and the commutativity of $\mathcal{A}$). Now, if $\alpha
\in\mathfrak{g}$, then $e^{\ast\alpha}=\sum\limits_{n\geq0}\dfrac{\alpha^{\ast
n}}{n!}$, so that%
\begin{align*}
e^{\ast\alpha}\circ Y  &  =\sum\limits_{n\geq0}\dfrac{\alpha^{\ast n}}%
{n!}\circ Y=\underbrace{\dfrac{\alpha^{\ast0}}{0!}}_{=e}\circ Y+\sum
\limits_{n\geq1}\dfrac{\alpha^{\ast n}}{n!}\circ Y=\underbrace{e\circ Y}%
_{=0}+\sum\limits_{n\geq1}\dfrac{\alpha^{\ast n}}{n!}\circ Y=\sum
\limits_{n\geq1}\dfrac{\alpha^{\ast n}}{n!}\circ Y\\
&  =\sum\limits_{n\geq1}\dfrac{1}{n!}\underbrace{\alpha^{\ast n}\circ
Y}_{=n\left(  \alpha\circ Y\right)  \ast\alpha^{\ast\left(  n-1\right)  }%
}=\sum\limits_{n\geq1}\underbrace{\dfrac{1}{n!}n}_{=\dfrac{1}{\left(
n-1\right)  !}}\left(  \alpha\circ Y\right)  \ast\alpha^{\ast\left(
n-1\right)  }\\
&  =\sum\limits_{n\geq1}\dfrac{1}{\left(  n-1\right)  !}\left(  \alpha\circ
Y\right)  \ast\alpha^{\ast\left(  n-1\right)  }=\left(  \alpha\circ Y\right)
\ast\sum\limits_{n\geq1}\dfrac{\alpha^{\ast\left(  n-1\right)  }}{\left(
n-1\right)  !}=\left(  \alpha\circ Y\right)  \ast\underbrace{\sum
\limits_{n\geq0}\dfrac{\alpha^{\ast n}}{n!}}_{=e^{\ast\alpha}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we substituted }n\text{ for
}n-1\text{ in the sum}\right) \\
&  =\left(  \alpha\circ Y\right)  \ast e^{\ast\alpha}=e^{\ast\alpha}%
\ast\left(  \alpha\circ Y\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by the
commutativity of }\mathcal{L}\left(  \mathcal{H},\mathcal{A}\right)  \right)
.
\end{align*}
But sinec $R\left(  \alpha\right)  $ is defined as the $\gamma\in\mathfrak{g}$
satisfying $e^{\ast\alpha}\circ Y=e^{\ast\alpha}\ast\gamma$, this yields
$R\left(  \alpha\right)  =\alpha\circ Y$, qed.\newline This proof works for
any field $k$ of characteristic $0$ (not necessarily $\mathbb{C}$) and for any
commutative $k$-algebra $\mathcal{A}$.

\item \textbf{IV.2:} On page 52, you write: "we define the Lie algebra :%
\[
\widetilde{g}=\mathfrak{g}\rtimes\mathbb{C}.Z_{0},
\]
". I think the $\widetilde{g}$ should be a $\widetilde{\mathfrak{g}}$ (since
you call it $\widetilde{\mathfrak{g}}$ later).

\item \textbf{IV.2:} On page 52, you write: "(see lemma II.7.1)". You probably
mean Lemma II.8.1 here.

\item \textbf{IV.2:} On page 52, you write: "We shall not dig out a Lie group
structure for $\widetilde{G}$ here" - but haven't you already dug it out? You
have defined the product in $\widetilde{G}$ in the sentence before, and I
assume the differentiable-manifold structure isn't much of a problem - or is it?

\item \textbf{IV.2:} At the very end of page 52, I don't understand Araki's
formula:%
\[
\exp\left(  tZ_{0}+\gamma\right)  =\sum_{n=0}^{\infty}\int_{\sum_{j=0}%
^{n}u_{j}=1,\ u_{j}\geq0}\exp\left(  u_{0}tZ_{0}\right)  \gamma\exp\left(
u_{1}tZ_{0}\right)  \gamma\cdots\gamma\exp\left(  u_{n}tZ_{0}\right)
du_{1}\cdots du_{n}.
\]
The $\exp\left(  u_{i}tZ_{0}\right)  $ terms lie in $\widetilde{G}$, while
$\gamma$ lies in $\mathfrak{g}$. How can we multiply them with each
other?\newline(I assume you define such a product by extending the law
$\left(  \varphi,t\right)  \left(  \psi,s\right)  =\left(  \varphi\ast\left(
\psi\circ\theta_{t}\right)  ,t+s\right)  $ to arbitrary $\varphi,\psi
\in\mathcal{L}\left(  \mathcal{H},\mathcal{A}\right)  $, rather than only
$\varphi,\psi\in G$. Can you confirm this?)

\item \textbf{IV.2:} On page 53, replace $x=\mathcal{H}^{n_{0}}$ by
$x\in\mathcal{H}^{n_{0}}$.

\item \textbf{IV.2:} On page 53, in the long calculation (the one that proves
the one-parameter group property), there are some typos:\newline- The second
$\sum\limits_{p=0}^{\infty}$ should be a $\sum\limits_{q=0}^{\infty}$%
.\newline- The very last integral should be $\int_{0\leq u_{n}\leq\cdots\leq
u_{1}\leq t+s}$ rather than $\int_{s\leq u_{p}\leq\cdots\leq u_{1}\leq t+s}$.

\item \textbf{Proof of Theorem IV.2.1:} I don't really understand the proof of
part 4) of this theorem at the place where you write "and thus replace the
group $G$ by any of the groups $G_{1},G_{2}$ in assertions 1), 2) and 3)".
This does seem to work for $G_{2}$, but I am not convinced that this works for
$G_{1}$: in fact, it is not clear how the argument that "The right-hand side
belongs manifestly to $G$" (this argument is on page 54, in the proof of part
1)) is supposed to work for $G_{1}$ (the sum of elements of $G_{1}$ needs not
be in $G_{1}$).\newline Anyway, I have an alternative proof of part 4), which
is devoid of any analysis and therefore works in the case of an arbitrary
commutative algebra $\mathcal{A}$ over any field $k$ of characteristic $0$
(not necessarily $k=\mathbb{C}$):\newline\textit{Alternative proof of Theorem
IV.2.1 part 4):} Let $\gamma\in\mathfrak{g}$. According to the proof of part
3) of Theorem IV.2.1, the element $\widetilde{R}^{-1}\left(  \gamma\right)  $
is the limit of the sequence $\left(  \varphi_{n}\right)  $ defined by
$\varphi_{0}=e$ and $\varphi_{n+1}=e+T\left(  \varphi_{n}\right)  $ for all
$n\geq0$, where $T\left(  \psi\right)  $ means $\left(  \psi\ast\gamma\right)
\circ Y^{-1}$.\newline Now, let $\gamma\in\mathfrak{g}_{2}$. Then, we are
going to prove that $\varphi_{n}\in G_{2}$ for all $n\in\mathbb{N}$.\newline
We prove this by induction over $n$: The induction base (the case $n=0$) is
obvious ($\varphi_{0}=e$ is an algebra homomorphism and thus lies in $G_{1}%
$).\newline For the induction step, let $m\in\mathbb{N}$ be arbitrary. Assume
that $\varphi_{m}\in G_{2}$. We now must prove that $\varphi_{m+1}=e+T\left(
\varphi_{m}\right)  $.\newline Since the recursive definition of the sequence
$\left(  \varphi_{n}\right)  $ yields $\varphi_{m+1}=e+T\left(  \varphi
_{m}\right)  $.\newline Since $\varphi_{m}\in G_{2}$, we know that
$\varphi_{m}$ is a cocycle. Since $\gamma$ is a cocycle as well (because
$\gamma\in\mathfrak{g}_{2}$), Proposition II.4.1 part 1) yields that
$\varphi_{m}\ast\gamma$ is a cocycle as well. Thus, $\left(  \varphi_{m}%
\ast\gamma\right)  \circ Y^{-1}$ is a cocycle. (This is because of the
following \textit{lemma:} If $\eta$ is a cocycle from $\mathcal{H}$ to
$\mathcal{A}$, then $\eta\circ Y^{-1}$ is a cocycle from $\mathcal{H}$ to
$\mathcal{A}$ as well.\footnote{\textit{Proof of the lemma:} Let $x$ and $y$
be two homogeneous elements of $\mathcal{H}$ such that $\left\vert
x\right\vert +\left\vert y\right\vert >0$. Then, $Y^{-1}\left(  xy\right)
=\dfrac{1}{\left\vert xy\right\vert }xy=\dfrac{1}{\left\vert x\right\vert
+\left\vert y\right\vert }xy$ (since $\left\vert xy\right\vert =\left\vert
x\right\vert +\left\vert y\right\vert $) and
\[
\left(  \eta\circ Y^{-1}\right)  \left(  xy\right)  =\eta\left(
\underbrace{Y^{-1}\left(  xy\right)  }_{=\dfrac{1}{\left\vert x\right\vert
+\left\vert y\right\vert }xy}\right)  =\eta\left(  \dfrac{1}{\left\vert
x\right\vert +\left\vert y\right\vert }xy\right)  =\dfrac{1}{\left\vert
x\right\vert +\left\vert y\right\vert }\eta\left(  xy\right)  .
\]
Similarly, $\left(  \eta\circ Y^{-1}\right)  \left(  yx\right)  =\dfrac
{1}{\left\vert y\right\vert +\left\vert x\right\vert }\eta\left(  yx\right)
$. Comparing these two equalities (and recalling that $\eta\left(  xy\right)
=\eta\left(  yx\right)  $ (since $\eta$ is a cocyle) and $\left\vert
x\right\vert +\left\vert y\right\vert =\left\vert y\right\vert +\left\vert
x\right\vert $), we conclude that $\left(  \eta\circ Y^{-1}\right)  \left(
xy\right)  =\left(  \eta\circ Y^{-1}\right)  \left(  yx\right)  $. We have
thus shown the identity $\left(  \eta\circ Y^{-1}\right)  \left(  xy\right)
=\left(  \eta\circ Y^{-1}\right)  \left(  yx\right)  $ for any two homogeneous
elements $x$ and $y$ of $\mathcal{H}$ satisfying $\left\vert x\right\vert
+\left\vert y\right\vert >0$. Since the identity $\left(  \eta\circ
Y^{-1}\right)  \left(  xy\right)  =\left(  \eta\circ Y^{-1}\right)  \left(
yx\right)  $ also trivially holds for any two homogeneous elements $x$ and $y$
of $\mathcal{H}$ satisfying $\left\vert x\right\vert +\left\vert y\right\vert
=0$ (because $\left\vert x\right\vert =\left\vert y\right\vert =0$ yields
$x,y\in k\cdot\mathbf{1}$), this yields that the identity $\left(  \eta\circ
Y^{-1}\right)  \left(  xy\right)  =\left(  \eta\circ Y^{-1}\right)  \left(
yx\right)  $ holds for any two homogeneous elements $x$ and $y$ of
$\mathcal{H}$. Since this identity is linear in $x$ and $y$, we can thus
conclude that this identity holds for \textit{any two} (not necessarily
homogeneous) elements $x$ and $y$ of $\mathcal{H}$. In other words, $\eta\circ
Y$ is a cocyle. The lemma is proven.})\newline By the definition of $T$, we
have $T\left(  \varphi_{m}\right)  =\left(  \varphi_{m}\ast\gamma\right)
\circ Y^{-1}$. Since $\left(  \varphi_{m}\ast\gamma\right)  \circ Y^{-1}$ is a
cocycle, this shows that $T\left(  \varphi_{m}\right)  $ is a cocycle. Since
$e$ is a cocycle, this yields that $e+T\left(  \varphi_{m}\right)  $ is a
cocycle (being the sum of two cocycles). Since $\left(  e+T\left(  \varphi
_{m}\right)  \right)  \left(  \mathbf{1}\right)  =\mathbf{1}_{\mathcal{A}}$,
this entails $e+T\left(  \varphi_{m}\right)  \in G_{2}$. We thus have
$\varphi_{m+1}=e+T\left(  \varphi_{m}\right)  \in G_{2}$.\newline This
completes the induction, and thus we have proven that $\varphi_{n}\in G_{2}$
for all $n\in\mathbb{N}$. Thus, the limit of the sequence $\left(  \varphi
_{n}\right)  $ is in $G_{2}$ as well (since the limit of a convergent sequence
of elements of $G_{2}$ must always be an element of $G_{2}$). In other words,
$\widetilde{R}^{-1}\left(  \gamma\right)  $ is in $G_{2}$ (since
$\widetilde{R}^{-1}\left(  \gamma\right)  $ is the limit of the sequence
$\left(  \varphi_{n}\right)  $).\newline We have thus shown that
$\widetilde{R}^{-1}\left(  \gamma\right)  \in G_{2}$ for every $\gamma
\in\mathfrak{g}_{2}$. In other words, we have shown that $\widetilde{R}%
^{-1}\left(  \mathfrak{g}_{2}\right)  \subseteq G_{2}$.\newline Now let us
prove that $\widetilde{R}^{-1}\left(  \mathfrak{g}_{1}\right)  \subseteq
G_{1}$:\newline Forget about the $\gamma\in\mathfrak{g}_{2}$ we took above.
Now let us take some $\gamma\in\mathfrak{g}_{1}$. Then, $\gamma$ is a
derivation from $\mathcal{H}$ with values in $\mathcal{A}$ and satisfies
$\gamma\left(  \mathbf{1}\right)  =0$.\newline Let $\varphi=\widetilde{R}%
^{-1}\left(  \gamma\right)  $. By the definition of $\widetilde{R}$, this
means that $\varphi\in G$ and $\varphi\circ Y=\varphi\ast\gamma$. From
$\varphi\in G$, we conclude that $\varphi\left(  \mathbf{1}\right)
=\mathbf{1}_{\mathcal{A}}$.\newline Every homogeneous $x\in\mathcal{H}$
satisfies%
\[
\left(  \varphi\circ Y\right)  \left(  x\right)  =\varphi\left(
\underbrace{Y\left(  x\right)  }_{=\left\vert x\right\vert x}\right)
=\varphi\left(  \left\vert x\right\vert x\right)  =\left\vert x\right\vert
\varphi\left(  x\right)  ,
\]
so that%
\begin{equation}
\left\vert x\right\vert \varphi\left(  x\right)  =\underbrace{\left(
\varphi\circ Y\right)  }_{=\varphi\ast\gamma}\left(  x\right)  =\left(
\varphi\ast\gamma\right)  \left(  x\right)  =\sum\limits_{\left(  x\right)
}\varphi\left(  x_{1}\right)  \gamma\left(  x_{2}\right)  . \label{iv0}%
\end{equation}
\newline We are now going to prove that%
\begin{equation}
\varphi\left(  xy\right)  =\varphi\left(  x\right)  \varphi\left(  y\right)
\ \ \ \ \ \ \ \ \ \ \text{for any two homogeneous elements }x\text{ and
}y\text{ of }\mathcal{H}. \label{iv1}%
\end{equation}
\newline\textit{Proof of (\ref{iv1}).} We will prove (\ref{iv1}) by strong
induction over $\left\vert x\right\vert +\left\vert y\right\vert $:\newline
The induction base (the case $\left\vert x\right\vert +\left\vert y\right\vert
=0$) is clear (since $\mathcal{H}_{0}=k\cdot\mathbf{1}$ and $\varphi\left(
\mathbf{1}\right)  =\mathbf{1}_{\mathcal{A}}$).\newline Now to the induction
step: Let $d\in\mathbb{N}$ be positive. Assume that (\ref{iv1}) is proven for
any two homogeneous elements $x$ and $y$ of $\mathcal{H}$ satisfying
$\left\vert x\right\vert +\left\vert y\right\vert <d$. Now let us prove
(\ref{iv1}) for any two homogeneous elements $x$ and $y$ of $\mathcal{H}$
satisfying $\left\vert x\right\vert +\left\vert y\right\vert =d$.\newline Let
$x$ and $y$ be two homogeneous elements of $\mathcal{H}$ satisfying
$\left\vert x\right\vert +\left\vert y\right\vert =d$. Then, we can WLOG
assume that $x=\sum\limits_{\left(  x\right)  }x_{1}\otimes x_{2}$ with
homogeneous $x_{1}$ and $x_{2}$ satisfying $\left\vert x_{1}\right\vert
+\left\vert x_{2}\right\vert =\left\vert x\right\vert $, and that
$y=\sum\limits_{\left(  y\right)  }y_{1}\otimes y_{2}$ with homogeneous
$y_{1}$ and $y_{2}$ satisfying $\left\vert y_{1}\right\vert +\left\vert
y_{2}\right\vert =y$.\newline Any three homogeneous elements $u$, $v$ and $w$
of $\mathcal{H}$ satisfying $\left\vert u\right\vert +\left\vert v\right\vert
+\left\vert w\right\vert =d$ satisfy%
\begin{equation}
\varphi\left(  uv\right)  \gamma\left(  w\right)  =\varphi\left(  u\right)
\varphi\left(  v\right)  \gamma\left(  w\right)  . \label{iv2}%
\end{equation}
\footnote{\textit{Proof of (\ref{iv2}).} We distinguish between two
cases:\newline\textit{Case 1:} We have $\left\vert w\right\vert =0$%
.\newline\textit{Case 2:} We have $\left\vert w\right\vert >0$.\newline In
Case 1, we have $\gamma\left(  w\right)  =0$ (since $\left\vert w\right\vert
=0$ yields $w\in\mathcal{H}^{0}=k\cdot\mathbf{1}$, and since we know that
$\gamma\left(  \mathbf{1}\right)  =0$). Therefore, (\ref{iv2}) trivially holds
in Case 1.\newline Let us now consider Case 2. In this case, $\left\vert
u\right\vert +\left\vert v\right\vert +\left\vert w\right\vert =d$ rewrites as
$\left\vert u\right\vert +\left\vert v\right\vert =d-\underbrace{\left\vert
w\right\vert }_{>0}<d$. Thus, in Case 2, we can apply (\ref{iv1}) to $u$ and
$v$ instead of $x$ and $y$ (since we assumed that (\ref{iv1}) is proven for
any two homogeneous elements $x$ and $y$ of $\mathcal{H}$ satisfying
$\left\vert x\right\vert +\left\vert y\right\vert <d$), and we obtain
$\varphi\left(  uv\right)  =\varphi\left(  u\right)  \varphi\left(  v\right)
$. This shows that (\ref{iv2}) holds in Case 2.\newline We have now shown that
(\ref{iv2}) holds in each of the two cases 1 and 2. This completes the proof
of (\ref{iv2}).} But (\ref{iv0}) (applied to $xy$ instead of $x$) yields%
\begin{align*}
\left\vert xy\right\vert \varphi\left(  xy\right)   &  =\sum\limits_{\left(
xy\right)  }\varphi\left(  \left(  xy\right)  _{1}\right)  \gamma\left(
\left(  xy\right)  _{2}\right)  =\sum\limits_{\left(  x\right)  \left(
y\right)  }\underbrace{\varphi\left(  x_{1}y_{1}\right)  \gamma\left(
x_{2}y_{2}\right)  }_{\substack{=\varphi\left(  x_{1}\right)  \varphi\left(
y_{1}\right)  \gamma\left(  x_{2}y_{2}\right)  \\\text{(by (\ref{iv2}),
applied to }u=x_{1}\text{,}\\v=y_{1}\text{ and }w=x_{2}y_{2}\text{)}}%
}=\sum\limits_{\left(  x\right)  \left(  y\right)  }\varphi\left(
x_{1}\right)  \varphi\left(  y_{1}\right)  \underbrace{\gamma\left(
x_{2}y_{2}\right)  }_{\substack{=e\left(  x_{2}\right)  \gamma\left(
y_{2}\right)  +\gamma\left(  x_{2}\right)  e\left(  y_{2}\right)
\\\text{(since }\gamma\text{ is a derivation)}}}\\
&  =\sum\limits_{\left(  x\right)  \left(  y\right)  }\varphi\left(
x_{1}\right)  \varphi\left(  y_{1}\right)  \left(  e\left(  x_{2}\right)
\gamma\left(  y_{2}\right)  +\gamma\left(  x_{2}\right)  e\left(
y_{2}\right)  \right) \\
&  =\underbrace{\sum\limits_{\left(  x\right)  \left(  y\right)  }%
\varphi\left(  x_{1}\right)  \varphi\left(  y_{1}\right)  e\left(
x_{2}\right)  \gamma\left(  y_{2}\right)  }_{\substack{=\sum\limits_{\left(
x\right)  }\varphi\left(  x_{1}\right)  e\left(  x_{2}\right)  \sum
\limits_{\left(  y\right)  }\varphi\left(  y_{1}\right)  \gamma\left(
y_{2}\right)  \\\text{(since }\mathcal{A}\text{ is commutative)}%
}}+\underbrace{\sum\limits_{\left(  x\right)  \left(  y\right)  }%
\varphi\left(  x_{1}\right)  \varphi\left(  y_{1}\right)  \gamma\left(
x_{2}\right)  e\left(  y_{2}\right)  }_{\substack{=\sum\limits_{\left(
x\right)  }\varphi\left(  x_{1}\right)  \gamma\left(  x_{2}\right)
\sum\limits_{\left(  y\right)  }\varphi\left(  y_{1}\right)  e\left(
y_{2}\right)  \\\text{(since }\mathcal{A}\text{ is commutative)}}}\\
&  =\underbrace{\sum\limits_{\left(  x\right)  }\varphi\left(  x_{1}\right)
e\left(  x_{2}\right)  }_{=\left(  \varphi\ast e\right)  \left(  x\right)
}\underbrace{\sum\limits_{\left(  y\right)  }\varphi\left(  y_{1}\right)
\gamma\left(  y_{2}\right)  }_{=\left(  \varphi\ast\gamma\right)  \left(
y\right)  }+\underbrace{\sum\limits_{\left(  x\right)  }\varphi\left(
x_{1}\right)  \gamma\left(  x_{2}\right)  }_{=\left(  \varphi\ast
\gamma\right)  \left(  x\right)  }\underbrace{\sum\limits_{\left(  y\right)
}\varphi\left(  y_{1}\right)  e\left(  y_{2}\right)  }_{=\left(  \varphi\ast
e\right)  \left(  x\right)  }\\
&  =\underbrace{\left(  \varphi\ast e\right)  }_{=\varphi}\left(  x\right)
\cdot\underbrace{\left(  \varphi\ast\gamma\right)  }_{=\varphi\circ Y}\left(
y\right)  +\underbrace{\left(  \varphi\ast\gamma\right)  }_{=\varphi\circ
Y}\left(  x\right)  \cdot\underbrace{\left(  \varphi\ast e\right)  }%
_{=\varphi}\left(  y\right) \\
&  =\varphi\left(  x\right)  \cdot\underbrace{\left(  \varphi\circ Y\right)
\left(  y\right)  }_{=\varphi\left(  Y\left(  y\right)  \right)
}+\underbrace{\left(  \varphi\circ Y\right)  \left(  x\right)  }%
_{=\varphi\left(  Y\left(  x\right)  \right)  }\cdot\varphi\left(  y\right) \\
&  =\varphi\left(  x\right)  \cdot\varphi\left(  \underbrace{Y\left(
y\right)  }_{=\left\vert y\right\vert y}\right)  +\varphi\left(
\underbrace{Y\left(  x\right)  }_{=\left\vert x\right\vert x}\right)
\cdot\varphi\left(  y\right) \\
&  =\varphi\left(  x\right)  \cdot\underbrace{\varphi\left(  \left\vert
y\right\vert y\right)  }_{=\left\vert y\right\vert \varphi\left(  y\right)
}+\underbrace{\varphi\left(  \left\vert x\right\vert x\right)  }_{=\left\vert
x\right\vert \varphi\left(  x\right)  }\cdot\varphi\left(  y\right)
=\left\vert y\right\vert \varphi\left(  x\right)  \cdot\varphi\left(
y\right)  +\left\vert x\right\vert \varphi\left(  x\right)  \cdot
\varphi\left(  y\right) \\
&  =\underbrace{\left(  \left\vert y\right\vert +\left\vert x\right\vert
\right)  }_{=\left\vert xy\right\vert }\varphi\left(  x\right)  \varphi\left(
y\right)  =\left\vert xy\right\vert \varphi\left(  x\right)  \varphi\left(
y\right)  .
\end{align*}
Since $\left\vert xy\right\vert =\left\vert x\right\vert +\left\vert
y\right\vert =d>0$, we can divide this equation by $\left\vert xy\right\vert $
and obtain $\varphi\left(  xy\right)  =\varphi\left(  x\right)  \varphi\left(
y\right)  $. In other words, we have proven (\ref{iv1}) for any two
homogeneous elements $x$ and $y$ of $\mathcal{H}$ satisfying $\left\vert
x\right\vert +\left\vert y\right\vert =d$. This completes the induction
step.\newline Thus, (\ref{iv1}) is proven for any two homogeneous elements $x$
and $y$ of $\mathcal{H}$. Since this equation (\ref{iv1}) is linear in $x$ and
$y$, we thus conclude that (\ref{iv1}) holds for any two elements $x$ and $y$
of $\mathcal{H}$ (not necessarily homogeneous). In other words, $\varphi$ is
an algebra morphism (since we also know that $\varphi\left(  \mathbf{1}%
\right)  =\mathbf{1}_{\mathcal{A}}$). In other words, $\varphi\in G_{1}$.
Thus, $\widetilde{R}^{-1}\left(  \gamma\right)  =\varphi\in G_{1}$.\newline We
have thus shown that $\widetilde{R}^{-1}\left(  \gamma\right)  \in G_{1}$ for
every $\gamma\in\mathfrak{g}_{1}$. Hence, $\widetilde{R}^{-1}\left(
\mathfrak{g}_{1}\right)  \subseteq G_{1}$. Combined with $\widetilde{R}%
^{-1}\left(  \mathfrak{g}_{2}\right)  \subseteq G_{2}$, this completes our
proof of part 4) of Proposition IV.2.1.

\item \textbf{Proof of Lemma IV.4.2:} In the first line of the computation
that you do in this proof, replace $\psi\circ Y^{n}$ by $z^{n}\psi\circ Y^{n}$.

\item \textbf{Proof of Theorem IV.4.1:} On page 57, you write: "We have still
to fix the convergence of the exponential just above in the case when
$z\widetilde{R}\left(  \psi\right)  $ belongs to $L\left(  \mathcal{H}%
,\mathcal{A}_{+}\right)  $." The $L$ here should be a calligraphic
$\mathcal{L}$.

\item \textbf{Proof of Theorem IV.4.1:} On page 57, you write:%
\[
\mathcal{L}_{+}^{n}=\bigcup_{p+q=n}\mathcal{L}_{+}^{p,q}.
\]
I assume the $\bigcup$ sign should be a $\sum$.

\item \textbf{Proof of Theorem IV.4.1:} On page 58, you write:%
\[
\dfrac{d}{dt}\mid_{t=0}\psi_{t}=z\left(  \psi\circ Y\right)  =\psi\dot{h}%
_{t}\mid_{t=0}.
\]
The $\psi_{t}$ should be a $\psi^{t}$ here.

\item \textbf{Theorem IV.4.4:} When you say $\mathbb{C}$ in this theorem, you
mean $\mathbb{C}$ as a subspace of $\mathcal{A}$. Maybe it would be good to
state this explicitly (I got confused by this $\mathbb{C}$).

\item \textbf{Proof of Lemma IV.4.5:} The reference to "theorem II.4.1" should
refer to Theorem II.5.1 instead.

\item \textbf{Proof of Theorem IV.4.4:} On page 60, you write: "it is easily
seen by induction on $\left\vert x\right\vert $ that the right-hand side
evaluated at $z$ has a limit when $z$ tends to infinity. Thus $\psi\left(
x\right)  \in\mathcal{A}_{-}$". I am not sure about the "Thus" here: the
function $\dfrac{1}{z-1}$ also has a limit when $z$ tends to infinity, but
doesn't lie in $\mathcal{A}_{-}$ for $z_{0}=0$. on the other hand, I think you
don't need the detour through limits: you can just say that "it is thus easily
seen by induction on $\left\vert x\right\vert $ that $\psi\left(  x\right)
\in\mathcal{A}_{-}$". Or am I missing something here?
\end{itemize}


\end{document}