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\begin{center}
\textbf{Higher Lie idempotents}

\textit{Fr\'{e}d\'{e}ric Patras and Christophe Reutenauer}

http://lacim.uqam.ca/\symbol{126}christo/idempotents.pdf (version November 18, 1998)

\textbf{Errata and questions - I (version 2)}

\bigskip
\end{center}

\begin{itemize}
\item \textbf{Page 1:} Typo: "caracteristic" should be "characteristic".

\item \textbf{Pages 1 and 2:} Typo: "envelopping" should be "enveloping" (this
typo appears several times).

\item \textbf{Page 2 and further:} Typo: "familly" should be "family" (this
typo appears several times).

\item \textbf{Page 2:} Maybe "Given a familly of Lie idempotents" should be
"Given an arbitrary Lie idempotent"? I think the constructions of the higher
Lie idempotents depend only on one Lie idempotent $\iota$ and (in the case of
higher Lie idempotents of the third kind) on a family of coefficients $a_{\mu
}^{\iota}$.

\item \textbf{Page 3:} Typo: "reodering" should be "reordering".

\item \textbf{Page 4:} Between Definition 2.2 and the Example, you write that
"the $\iota$-descent algebra decomposes as a direct sum%
\[
\mathcal{D}_{\iota}=\bigoplus_{n=0}^{\infty}\mathcal{D}_{\iota n}.
\]
". It might be useful to notice here that this is a direct sum of vector
spaces, not of algebras (under the convolution $\ast$).

\item \textbf{Page 5:} In the proof of Lemma 3.1, you write: "More generally,
for any $l\geq2$ and $k\geq3$, let $\Delta_{l2}$ be [...]". I don't see any
reason to require $l\geq2$ and $k\geq3$ here; everything is just as correct
for any $l\geq0$ and $m\geq0$.

\item \textbf{Page 6:} In the proof of Lemma 3.1, the $\sum_{\sigma\in S_{n}}$
should be $\sum_{\sigma\in S_{k}}$.

\item \textbf{Page 6:} In the proof of Lemma 3.1, you write: "If we apply
$\Pi_{k}$ to the whole sum" (in the fourth line of page 6). I think you are
applying $\Pi_{k}^{\otimes k}$ here, not $\Pi_{k}$.

\item \textbf{Page 6:} In the proof of Lemma 3.1, you have a typo: "Aplying"
should be "Applying".

\item \textbf{Page 6:} In the proof of Lemma 3.1, you write: "Now,this sum is
equal to $\sum\left(  \iota_{\mu_{1}}\otimes...\otimes\iota_{\mu_{k}}\right)
\circ\sigma\left(  x_{1}\otimes...\otimes x_{k}\right)  $, where $\sigma$
denotes here the natural action of the symmetric group on $A^{\otimes n}$".
First, there should be a whitespace after "Now,". Second, the $\sigma\left(
x_{1}\otimes...\otimes x_{k}\right)  $ should be a $\sigma^{-1}\left(
x_{1}\otimes...\otimes x_{k}\right)  $, because $\sigma\left(  x_{1}%
\otimes...\otimes x_{k}\right)  $ is $x_{\sigma^{-1}\left(  1\right)  }%
\otimes...\otimes x_{\sigma^{-1}\left(  k\right)  }$ rather than
$x_{\sigma\left(  1\right)  }\otimes...\otimes x_{\sigma\left(  k\right)  }$.
Third, I think you mean $A^{\otimes k}$ instead of $A^{\otimes n}$ (unless you
want to talk about general $n$).

\item \textbf{Page 6:} In the proof of Lemma 3.1, you write: "Since the
coproduct is cocommutative, we deduce that
\[
\left(  \iota_{\mu_{1}}\otimes...\otimes\iota_{\mu_{k}}\right)  \circ\left(
\Pi_{k}^{\otimes k}\right)  \circ\Delta_{kk}\circ\left(  \iota_{\lambda_{1}%
}\otimes...\otimes\iota_{\lambda_{k}}\right)  =\sum\left(  \iota_{\mu_{1}%
}\otimes...\otimes\iota_{\mu_{k}}\right)  \circ\sigma\circ\Delta_{k}%
=\sum\left(  \iota_{\mu_{1}}\otimes...\otimes\iota_{\mu_{k}}\right)
\circ\Delta_{k},
\]
which implies (ii)." The $\sigma$ here should be a $\sigma^{-1}$. (Also, what
somewhat confused me is that cocommutativity is used in the passage from
$\sum\left(  \iota_{\mu_{1}}\otimes...\otimes\iota_{\mu_{k}}\right)
\circ\sigma\circ\Delta_{k}$ to $\sum\left(  \iota_{\mu_{1}}\otimes
...\otimes\iota_{\mu_{k}}\right)  \circ\Delta_{k}$, not in the passage from
$\left(  \iota_{\mu_{1}}\otimes...\otimes\iota_{\mu_{k}}\right)  \circ\left(
\Pi_{k}^{\otimes k}\right)  \circ\Delta_{kk}\circ\left(  \iota_{\lambda_{1}%
}\otimes...\otimes\iota_{\lambda_{k}}\right)  $ to $\sum\left(  \iota_{\mu
_{1}}\otimes...\otimes\iota_{\mu_{k}}\right)  \circ\sigma\circ\Delta_{k}$. It
thus would probably better to mention cocommutativity after the long equation
rather than before it.)

\item \textbf{Page 7:} In the proof of Theorem 3.4, you write: "Thus $f$ is
idempotent if and only if [...]". But in general, only the "if" part of this
is true (and fortunately, only the "if" part is needed), since nobody has told
us that the $\iota_{\alpha}$ are linearly independent.

\item \textbf{Page 7:} In the proof of Theorem 3.4, it would be clearer if you
replace $\left(  n_{1}+...+n_{k}\right)  !/n_{1}!...n_{k}!$ by $\left(
n_{1}+...+n_{k}\right)  !/\left(  n_{1}!...n_{k}!\right)  $. (I consider the
notation $a/b_{1}b_{2}...b_{k}$ for $a/\left(  b_{1}b_{2}...b_{k}\right)  $
outdated and ambiguous, although it seems to be still in use.)

\item \textbf{Page 7:} In Definition 4.1, I feel it would be good to point out
three things explicitly:

-- The "$1$" in "$F_{\lambda}^{\iota}:=\left(  1-\sum\limits_{l\left(
\mu\right)  <l\left(  \lambda\right)  }F_{\mu}^{\iota}\right)  \circ
E_{\lambda}^{\iota}$" means the identity map $\operatorname*{id}%
\nolimits_{A_{n}}\in\operatorname*{End}\left(  A_{n}\right)  $, not the unity
of the algebra $\mathcal{L}\left(  A\right)  $.

-- For $n=0$, the element $F_{\left(  {}\right)  }^{\iota}$ is defined as
$E_{\left(  {}\right)  }^{\iota}=\operatorname*{id}\nolimits_{A_{0}}=\eta
\circ\epsilon$ (here we are using the identification of $\operatorname*{End}%
\left(  A_{0}\right)  $ with the space of all graded endomorphisms of $A$
whose image is $\subseteq A_{0}$). (While this can be seen as a consequence of
the formula $F_{\lambda}^{\iota}:=\left(  1-\sum\limits_{l\left(  \mu\right)
<l\left(  \lambda\right)  }F_{\mu}^{\iota}\right)  \circ E_{\lambda}^{\iota}$
applied to $\lambda=\left(  {}\right)  $, it would be helpful to point this
out explicitly).

-- The maps $F_{\lambda}^{\iota}$ are called the "higher Lie idempotents of
the second kind".

\item \textbf{Page 7:} In Definition 4.1, it wouldn't harm to say that the
"induction base" $F_{\left(  n\right)  }^{\iota}:=E_{\left(  n\right)
}^{\iota}=\iota_{n}$ is, itself, a particular case of the "induction step"
$F_{\lambda}^{\iota}:=\left(  1-\sum\limits_{l\left(  \mu\right)  <l\left(
\lambda\right)  }F_{\mu}^{\iota}\right)  \circ E_{\lambda}^{\iota}$. In fact,
if we substitute $\lambda=\left(  n\right)  $ in $F_{\lambda}^{\iota}:=\left(
1-\sum\limits_{l\left(  \mu\right)  <l\left(  \lambda\right)  }F_{\mu}^{\iota
}\right)  \circ E_{\lambda}^{\iota}$, then we get $F_{\left(  n\right)
}^{\iota}=\left(  1-\sum\limits_{l\left(  \mu\right)  <l\left(  \left(
n\right)  \right)  }F_{\mu}^{\iota}\right)  \circ E_{\left(  n\right)
}^{\iota}$, but the sum $\sum\limits_{l\left(  \mu\right)  <l\left(  \left(
n\right)  \right)  }F_{\mu}^{\iota}$ is empty since $l\left(  \left(
n\right)  \right)  =1$, and thus this becomes $F_{\left(  n\right)  }^{\iota
}=E_{\left(  n\right)  }^{\iota}$.\newline This fact allows us to use
$F_{\lambda}^{\iota}=\left(  1-\sum\limits_{l\left(  \mu\right)  <l\left(
\lambda\right)  }F_{\mu}^{\iota}\right)  \circ E_{\lambda}^{\iota}$ not only
for $\lambda\neq\left(  n\right)  $ but also for all $\lambda$. This is used
in several proofs in your paper.

\item \textbf{Page 7:} In the Remark 1) at the end of page 7, you made a typo:
"othogonal" should be "orthogonal".

\item \textbf{Page 9:} On the first line of this page, you write: "$F_{\mu
}^{\iota}\circ F_{\beta}^{\iota}=\delta_{\mu\beta}$". This should be $F_{\mu
}^{\iota}\circ F_{\beta}^{\iota}=\delta_{\mu\beta}F_{\mu}^{\iota}$. (The only
thing you actually use, though, is that $F_{\mu}^{\iota}\circ F_{\beta}%
^{\iota}=0$ for $\mu\neq\beta$ when $l\left(  \mu\right)  $ and $l\left(
\beta\right)  $ are both $<k$.)

\item \textbf{Page 9:} In the proof of Theorem 4.3, you write: "we have by
Def.4.1 that $E_{\lambda}^{\iota}\left(  x\right)  =F_{\lambda}^{\iota}\left(
x\right)  $ plus a sum of $E_{\lambda_{1}}^{\iota}\circ...\circ E_{\lambda
_{k}}^{\iota}$". First, either you should replace the $E_{\lambda}^{\iota
}\left(  x\right)  $ and $F_{\lambda}^{\iota}\left(  x\right)  $ here by
$E_{\lambda}^{\iota}$ and $F_{\lambda}^{\iota}$, or you should replace the
$E_{\lambda_{1}}^{\iota}\circ...\circ E_{\lambda_{k}}^{\iota}$ by an $\left(
E_{\lambda_{1}}^{\iota}\circ...\circ E_{\lambda_{k}}^{\iota}\right)  \left(
x\right)  $. Second, "sum" is slightly imprecise; you mean a linear
combination rather than a sum (the coefficients in this combination can be
both $+1$ and $-1$).

\item \textbf{Page 9:} In the proof of Theorem 4.3, you write: "the elements
$\left(  a_{1},...,a_{k}\right)  =\left(  1/k!\right)  \sum_{k\in S_{k}%
}a_{\sigma\left(  1\right)  }...a_{\sigma\left(  k\right)  }$". Replace
$\sum_{k\in S_{k}}$ by $\sum_{\sigma\in S_{k}}$ here.

\item \textbf{Page 9:} In the proof of Theorem 4.3, you write:\newline"Since
$A$ is a graded cocommutative connected bialgebra of characteristic zero, it
is by the Cartier-Milnor-Moore theorem isomorphic to the envelopping algebra
of $Prim\left(  A\right)  $. Hence, by the Poincar\'{e}-Birkhoff-Witt theorem
it is the direct sum of its subspaces $A^{\lambda}$, where for any partition
$\lambda$, the latter subspace is spanned by the elements $\left(
a_{1},...,a_{k}\right)  =\left(  1/k!\right)  \sum_{k\in S_{k}}a_{\sigma
\left(  1\right)  }...a_{\sigma\left(  k\right)  }$, for any choice of
homogeneous primitive elements $a_{i}$, with $\deg\left(  a_{i}\right)
=\lambda_{i}$ and $\lambda=\left(  \lambda_{1},...,\lambda_{k}\right)
$."\newline This is a correct argument (up to the typos I mentioned above),
but somewhat an overkill. In fact, you only need the easy part of the
Cartier-Milnor-Moore theorem\footnote{By the "easy part", I mean the statement
that a graded cocommutative connected bialgebra over a field of characteristic
$0$ is always generated as an algebra by its primitive elements.} and only the
easy part of the Poincar\'{e}-Birkhoff-Witt theorem\footnote{Here, the "easy
part" is the statement that the symmetrization map $S\left(  \mathfrak{g}%
\right)  \rightarrow U\left(  \mathfrak{g}\right)  $ is surjective. (This only
makes sense in characteristic $0$.)} to show that $A$ is the sum of its
subspaces $A^{\lambda}$ (we don't yet know that it is the \textit{direct}
sum), and this is already enough for your proof of Theorem 4.3. (I can detail
this argument better if you wish, but I have a feeling that you already know
this). Maybe you need something stronger (like the direct sum assertion) to
prove Corollary 4.4 though (I don't understand your proof at the moment), but
I would always try to do without - maybe this will net us an explicit
constructive proof of Poincar\'{e}-Birkhoff-Witt or Cartier-Milnor-Moore at
the end...

\item \textbf{Page 9:} In the proof of Theorem 4.3, you write: "It is equal to
$\sum_{\mu}\Pi_{k}\circ\iota_{\mu_{1}}\otimes...\otimes\iota_{\mu_{k}}%
\circ\Delta_{k}\left(  a_{1}...a_{k}\right)  $". I would put the $\iota
_{\mu_{1}}\otimes...\otimes\iota_{\mu_{k}}$ term in brackets here.

\item \textbf{Page 9:} In the last absatz of page 9, you write: "the
\textit{cofree cocommutative coalgebra} on a vector space $V$". But I think it
is more common to say "over a vector space $V$" rather than "on a vector space
$V$". (You yourself say "over" in Corollary 4.4.)

\item \textbf{Page 10:} In Corollary 4.4, replace "$\bigoplus\limits_{n\in
\mathbf{N}}\iota^{\otimes n}\circ\Delta_{n}$" by "$\bigoplus\limits_{n\in
\mathbf{N}}\dfrac{1}{n!}\iota^{\otimes n}\circ\Delta_{n}$" (otherwise, this
map would not be a coalgebra homomorphism).

\item \textbf{Page 10:} In Corollary 4.4, replace the $\mapsto$ arrow by a
$\rightarrow$ arrow.

\item \textbf{Page 10:} In Corollary 4.4, replace "$\dfrac{1}{l\left(
\lambda\right)  !}\left(  1-\sum\limits_{l\left(  \mu\right)  <l\left(
\lambda\right)  }F_{\mu}^{\iota}\right)  \circ\Pi_{k}$" by "$\left(
1-\sum\limits_{l\left(  \mu\right)  <l\left(  \lambda\right)  }F_{\mu}^{\iota
}\right)  \circ\Pi_{k}$" (this change is needed to "balance out" the
$\dfrac{1}{n!}$ factor I added to "$\bigoplus\limits_{n\in\mathbf{N}}%
\iota^{\otimes n}\circ\Delta_{n}$").

\item \textbf{Page 10:} You write that "The corollary follows, once it is
noted that $Sym^{\lambda}\left(  Prim\left(  A\right)  \right)  $ is
canonically isomorphic to $A^{\lambda}$, through the map $\Pi_{k}$". I do
understand why $\operatorname*{Sym}\nolimits^{\lambda}\left(
\operatorname*{Prim}A\right)  $ is canonically isomorphic to $A^{\lambda}$
through the map $\Pi_{k}$\ \ \ \ \footnote{In fact, let $\lambda=\left(
\lambda_{1},\lambda_{2},...,\lambda_{k}\right)  $. For every $F$-vector space
$V$ and every subset $S$ of $V$, let $\left\langle S\right\rangle $ denote the
$F$-linear span of the set $S$.
\par
By the definition of $\operatorname*{Sym}\nolimits^{\lambda}\left(
\operatorname*{Prim}A\right)  $, we know that $\operatorname*{Sym}%
\nolimits^{\lambda}\left(  \operatorname*{Prim}A\right)  $ is the $F$-linear
span of the elements $\dfrac{1}{k!}\sum\limits_{\sigma\in S_{k}}%
x_{\sigma\left(  1\right)  }\otimes x_{\sigma\left(  2\right)  }%
\otimes...\otimes x_{\sigma\left(  k\right)  }$ where $\left(  x_{1}%
,x_{2},...,x_{k}\right)  $ ranges over all $k$-tuples of homogeneous elements
of $\operatorname*{Prim}A$ satisfying $\left(  \deg\left(  x_{i}\right)
=\lambda_{i}\text{ for all }i\in\left\{  1,2,...,k\right\}  \right)  $. In
other words,%
\begin{align}
\operatorname*{Sym}\nolimits^{\lambda}\left(  \operatorname*{Prim}A\right)
&  =\left\langle \left\{  \dfrac{1}{k!}\sum\limits_{\sigma\in S_{k}}%
x_{\sigma\left(  1\right)  }\otimes x_{\sigma\left(  2\right)  }%
\otimes...\otimes x_{\sigma\left(  k\right)  }\ \mid\text{ all }x_{i}\text{
are homogeneous}\right.  \right.  \nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left.  \left.
\phantom{\sum\limits_{i\in S_k}}\text{\ elements of }\operatorname*{Prim}%
A\text{ and satisfy }\deg\left(  x_{i}\right)  =\lambda_{i}\text{ for all
}i\in\left\{  1,2,...,k\right\}  \right\}  \right\rangle \nonumber\\
&  =\left\langle \left\{  \dfrac{1}{k!}\sum\limits_{\sigma\in S_{k}}%
x_{\sigma\left(  1\right)  }\otimes x_{\sigma\left(  2\right)  }%
\otimes...\otimes x_{\sigma\left(  k\right)  }\ \mid\text{ all }x_{i}\text{
are primitive and homogeneous}\right.  \right.  \nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left.  \left.
\phantom{\sum\limits_{i\in S_k}}\text{\ elements of }A\text{ and satisfy }%
\deg\left(  x_{i}\right)  =\lambda_{i}\text{ for all }i\in\left\{
1,2,...,k\right\}  \right\}  \right\rangle \nonumber\\
&  =\left\langle \left\{  \dfrac{1}{k!}\sum\limits_{\sigma\in S_{k}}%
a_{\sigma\left(  1\right)  }\otimes a_{\sigma\left(  2\right)  }%
\otimes...\otimes a_{\sigma\left(  k\right)  }\ \mid\text{ all }a_{i}\text{
are primitive and homogeneous}\right.  \right.  \nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left.  \left.
\phantom{\sum\limits_{i\in S_k}}\text{\ elements of }A\text{ and satisfy }%
\deg\left(  a_{i}\right)  =\lambda_{i}\text{ for all }i\in\left\{
1,2,...,k\right\}  \right\}  \right\rangle \nonumber
\end{align}
(here, we renamed $x_{i}$ as $a_{i}$). In other words,%
\begin{align}
&  \left\langle \left\{  \dfrac{1}{k!}\sum\limits_{\sigma\in S_{k}}%
a_{\sigma\left(  1\right)  }\otimes a_{\sigma\left(  2\right)  }%
\otimes...\otimes a_{\sigma\left(  k\right)  }\ \mid\text{ all }a_{i}\text{
are primitive and homogeneous}\right.  \right.  \nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left.  \left.
\phantom{\sum\limits_{i\in S_k}}\text{\ elements of }A\text{ and satisfy }%
\deg\left(  a_{i}\right)  =\lambda_{i}\text{ for all }i\in\left\{
1,2,...,k\right\}  \right\}  \right\rangle \nonumber\\
&  =\operatorname*{Sym}\nolimits^{\lambda}\left(  \operatorname*{Prim}%
A\right)  .\tag{A1}%
\end{align}
\par
By the definition of $A^{\lambda}$, we know that $A^{\lambda}$ is the
$F$-linear span of the elements $\dfrac{1}{k!}\sum\limits_{\sigma\in S_{k}%
}a_{\sigma\left(  1\right)  }a_{\sigma\left(  2\right)  }...a_{\sigma\left(
k\right)  }$ where $\left(  a_{1},a_{2},...,a_{k}\right)  $ ranges over all
$k$-tuples of primitive homogeneous elements of $A$ satisfying $\left(
\deg\left(  a_{i}\right)  =\lambda_{i}\text{ for all }i\in\left\{
1,2,...,k\right\}  \right)  $. In other words,
\begin{align*}
A^{\lambda} &  =\left\langle \left\{  \dfrac{1}{k!}\sum\limits_{\sigma\in
S_{k}}a_{\sigma\left(  1\right)  }a_{\sigma\left(  2\right)  }...a_{\sigma
\left(  k\right)  }\ \mid\text{ all }a_{i}\text{ are primitive and
homogeneous}\right.  \right.  \\
&  \ \ \ \ \ \ \ \ \ \ \left.  \left.
\phantom{\sum\limits_{i\in S_k}}\text{\ elements of }A\text{ and satisfy }%
\deg\left(  a_{i}\right)  =\lambda_{i}\text{ for all }i\in\left\{
1,2,...,k\right\}  \right\}  \right\rangle \\
&  =\left\langle \left\{  \Pi_{k}\left(  \dfrac{1}{k!}\sum\limits_{\sigma\in
S_{k}}a_{\sigma\left(  1\right)  }\otimes a_{\sigma\left(  2\right)  }%
\otimes...\otimes a_{\sigma\left(  k\right)  }\right)  \ \mid\text{ all }%
a_{i}\text{ are primitive and homogeneous}\right.  \right.  \\
&  \ \ \ \ \ \ \ \ \ \ \left.  \left.
\phantom{\sum\limits_{i\in S_k}}\text{\ elements of }A\text{ and satisfy }%
\deg\left(  a_{i}\right)  =\lambda_{i}\text{ for all }i\in\left\{
1,2,...,k\right\}  \right\}  \right\rangle \\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{since }\dfrac{1}{k!}\sum\limits_{\sigma\in S_{k}}a_{\sigma\left(
1\right)  }a_{\sigma\left(  2\right)  }...a_{\sigma\left(  k\right)  }=\Pi
_{k}\left(  \dfrac{1}{k!}\sum\limits_{\sigma\in S_{k}}a_{\sigma\left(
1\right)  }\otimes a_{\sigma\left(  2\right)  }\otimes...\otimes
a_{\sigma\left(  k\right)  }\right)  \\
\text{for any }\left(  a_{1},a_{2},...,a_{k}\right)  \in A^{k}%
\end{array}
\right)  \\
&  =\left\langle \Pi_{k}\left(  \left\{  \dfrac{1}{k!}\sum\limits_{\sigma\in
S_{k}}a_{\sigma\left(  1\right)  }\otimes a_{\sigma\left(  2\right)  }%
\otimes...\otimes a_{\sigma\left(  k\right)  }\ \mid\text{ all }a_{i}\text{
are primitive and homogeneous}\right.  \right.  \right.  \\
&  \ \ \ \ \ \ \ \ \ \ \left.  \left.  \left.
\phantom{\sum\limits_{i\in S_k}}\text{\ elements of }A\text{ and satisfy }%
\deg\left(  a_{i}\right)  =\lambda_{i}\text{ for all }i\in\left\{
1,2,...,k\right\}  \right\}  \right)  \right\rangle \\
&  =\Pi_{k}\left(  \left\langle \left\{  \dfrac{1}{k!}\sum\limits_{\sigma\in
S_{k}}a_{\sigma\left(  1\right)  }\otimes a_{\sigma\left(  2\right)  }%
\otimes...\otimes a_{\sigma\left(  k\right)  }\ \mid\text{ all }a_{i}\text{
are primitive and homogeneous}\right.  \right.  \right)  \\
&  \ \ \ \ \ \ \ \ \ \ \left.  \left.  \left.
\phantom{\sum\limits_{i\in S_k}}\text{\ elements of }A\text{ and satisfy }%
\deg\left(  a_{i}\right)  =\lambda_{i}\text{ for all }i\in\left\{
1,2,...,k\right\}  \right\}  \right\rangle \right)  \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\Pi_{k}\text{ is }F\text{-linear}%
\right)  \\
&  =\Pi_{k}\left(  \operatorname*{Sym}\nolimits^{\lambda}\left(
\operatorname*{Prim}A\right)  \right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by
(A1)}\right)  .
\end{align*}
Hence, $\Pi_{k}$ restricts to a surjective homomorphism $\operatorname*{Sym}%
\nolimits^{\lambda}\left(  \operatorname*{Prim}A\right)  \rightarrow
A^{\lambda}$.
\par
Moreover, let $\widetilde{\Pi}$ be the homomorphism $\bigoplus\limits_{n\in
\mathbb{N}}\Pi_{n}\mid_{\left(  \left(  \operatorname*{Prim}A\right)
^{\otimes n}\right)  ^{S_{n}}}:\bigoplus\limits_{n\in\mathbb{N}}\left(
\left(  \operatorname*{Prim}A\right)  ^{\otimes n}\right)  ^{S_{n}}\rightarrow
A$ (composed of the homomorphisms $\Pi_{n}\mid_{\left(  \left(
\operatorname*{Prim}A\right)  ^{\otimes n}\right)  ^{S_{n}}}:\left(  \left(
\operatorname*{Prim}A\right)  ^{\otimes n}\right)  ^{S_{n}}\rightarrow A$ for
all $n\in\mathbb{N}$). This homomorphism $\widetilde{\Pi}$ sends $\dfrac
{1}{n!}\sum\limits_{\sigma\in S_{n}}a_{\sigma\left(  1\right)  }\otimes
a_{\sigma\left(  2\right)  }\otimes...\otimes a_{\sigma\left(  n\right)  }$ to
$\dfrac{1}{n!}\sum\limits_{\sigma\in S_{n}}a_{\sigma\left(  1\right)
}a_{\sigma\left(  2\right)  }...a_{\sigma\left(  n\right)  }$ for every
$n\in\mathbb{N}$ and $\left(  a_{1},a_{2},...,a_{n}\right)  \in\left(
\operatorname*{Prim}A\right)  ^{n}$. According to the
Poincar\'{e}-Birkhoff-Witt theorem, this homomorphism $\widetilde{\Pi}$ is an
isomorphism (since the Cartier-Milnor-Moore theorem yields $A\cong U\left(
\operatorname*{Prim}A\right)  $, and under the identification of $A$ with
$U\left(  \operatorname*{Prim}A\right)  $ the homomorphism $\widetilde{\Pi}$
becomes the symmetrization map $S\left(  \operatorname*{Prim}A\right)
\rightarrow U\left(  \operatorname*{Prim}A\right)  $). Hence, $\widetilde{\Pi
}$ is injective.
\par
Now, since $\operatorname*{Sym}\nolimits^{\lambda}\left(  \operatorname*{Prim}%
A\right)  \subseteq\left(  \left(  \operatorname*{Prim}A\right)  ^{\otimes
k}\right)  ^{S_{k}}$, we have%
\begin{align*}
\widetilde{\Pi}\mid_{\operatorname*{Sym}\nolimits^{\lambda}\left(
\operatorname*{Prim}A\right)  } &  =\underbrace{\left(  \widetilde{\Pi}%
\mid_{\left(  \left(  \operatorname*{Prim}A\right)  ^{\otimes k}\right)
^{S_{k}}}\right)  }_{\substack{=\Pi_{k}\mid_{\left(  \left(
\operatorname*{Prim}A\right)  ^{\otimes k}\right)  ^{S_{k}}}\\\text{(since
}\widetilde{\Pi}=\bigoplus\limits_{n\in\mathbb{N}}\Pi_{n}\mid_{\left(  \left(
\operatorname*{Prim}A\right)  ^{\otimes n}\right)  ^{S_{n}}}\text{)}}%
}\mid_{\operatorname*{Sym}\nolimits^{\lambda}\left(  \operatorname*{Prim}%
A\right)  }\\
&  =\left(  \Pi_{k}\mid_{\left(  \left(  \operatorname*{Prim}A\right)
^{\otimes k}\right)  ^{S_{k}}}\right)  \mid_{\operatorname*{Sym}%
\nolimits^{\lambda}\left(  \operatorname*{Prim}A\right)  }=\Pi_{k}%
\mid_{\operatorname*{Sym}\nolimits^{\lambda}\left(  \operatorname*{Prim}%
A\right)  }.
\end{align*}
Since $\widetilde{\Pi}\mid_{\operatorname*{Sym}\nolimits^{\lambda}\left(
\operatorname*{Prim}A\right)  }$ is injective (because $\widetilde{\Pi}$ is
injective), this yields that $\Pi_{k}\mid_{\operatorname*{Sym}%
\nolimits^{\lambda}\left(  \operatorname*{Prim}A\right)  }$ is injective.
\par
Now, consider the surjective homomorphism $\operatorname*{Sym}%
\nolimits^{\lambda}\left(  \operatorname*{Prim}A\right)  \rightarrow
A^{\lambda}$ to which $\Pi_{k}$ restricts. This homomorphism is also injective
(since $\Pi_{k}\mid_{\operatorname*{Sym}\nolimits^{\lambda}\left(
\operatorname*{Prim}A\right)  }$ is injective), and thus it is an isomorphism.
Thus, $\Pi_{k}$ restricts to an isomorphism $\operatorname*{Sym}%
\nolimits^{\lambda}\left(  \operatorname*{Prim}A\right)  \rightarrow
A^{\lambda}$. Hence, $\operatorname*{Sym}\nolimits^{\lambda}\left(
\operatorname*{Prim}A\right)  $ is isomorphic to $A^{\lambda}$ through the map
$\Pi_{k}$, qed.}. But I don't understand how Corollary 4.4 follows from this!
In particular, I don't see how the $\dfrac{1}{l\left(  \lambda\right)
!}\left(  1-\sum\limits_{l\left(  \mu\right)  <l\left(  \lambda\right)
}F_{\mu}^{\iota}\right)  $ term appears.

\item \textbf{Page 11:} In the proof of Theorem 5.1, you write: "We multiply
this by $e_{n}$ on the right in $\mathcal{L}\left(  A\right)  $". I think this
is confusing: Multiplying something in $\mathcal{L}\left(  A\right)  $ means
convolution, but you want composition. Maybe you could just say "We compose
this with $e_{n}$ on the right"?

\item \textbf{Page 11:} In the proof of Theorem 5.1, you write:\newline"Thus
we obtain $e_{n}=\alpha e_{n}$, since $e_{\mu}\circ e_{n}=0$ by Lemma 3.2.
Thus, in case $e_{n}\neq0$, $\alpha=1$; and in case $e_{n}=0$, we must have
also $\iota_{n}=0$, and we may take $\alpha=1$ in (*)."\newline This argument
is correct, but I think it can be simplified as follows:\newline"Thus we
obtain $e_{n}=\alpha e_{n}$, since $e_{\mu}\circ e_{n}=0$ by Lemma 3.2. Thus,
we can replace $\alpha e_{n}$ by $e_{n}$ in (*), and get $\iota_{n}=e_{n}%
+\sum_{\mu}\ast e_{\mu}$."\newline This simplified argument has the additional
advantage of being valid when $k$ is not necessarily a field.

\item \textbf{Page 11:} In the proof of Theorem 5.1, you made a typo: "matrix
fom" should be "matrix form".

\item \textbf{Page 11:} In the proof of Theorem 5.1, you write: "It is clear
that (i) implies (iv)". But is this really clear on its own, or is it clear
using the fact that $\mathcal{D}\left(  A\right)  $ is closed under
convolution (a consequence of Theorem 9.2 in [R2], but [R2] only considers the
case when $A$ is the tensor algebra of an alphabet)?

\item \textbf{Page 12:} In the proof of Lemma 5.3, replace $\iota_{\mu2}$ by
$\iota_{\mu_{2}}$ (you forgot to make the $2$ an index).

\item \textbf{Pages 12 and 13:} In the proof of Theorem 5.4, you write:
"Moreover:%
\[
\left(  \sum_{\mu<\left[  n\right]  }\mathcal{E}_{\mu}^{\iota}\right)
^{2}=\left(  1-\mathcal{E}_{\left[  n\right]  }^{\iota}\right)  ^{2}%
=1-\mathcal{E}_{\left[  n\right]  }^{\iota}=\sum_{\mu<\left[  n\right]
}\mathcal{E}_{\mu}^{\iota},
\]
and:%
\[
\left(  \sum_{\mu<\left[  n\right]  }\mathcal{E}_{\mu}^{\iota}\right)
\circ\mathcal{E}_{\left[  n\right]  }^{\iota}=\left(  1-\mathcal{E}_{\left[
n\right]  }^{\iota}\right)  \circ\mathcal{E}_{\left[  n\right]  }^{\iota}=0.
\]
"\newline These formulas are not literally true, because $\sum\limits_{\mu
<\left[  n\right]  }\mathcal{E}_{\mu}^{\iota}$ is $p_{n}-\mathcal{E}_{\left[
n\right]  }^{\iota}$ rather than $1-\mathcal{E}_{\left[  n\right]  }^{\iota}$
(since
\begin{align*}
\sum\limits_{\mu<\left[  n\right]  }\mathcal{E}_{\mu}^{\iota}+\mathcal{E}%
_{\left[  n\right]  }^{\iota} &  =\sum\limits_{\mu\leq\left[  n\right]
}\mathcal{E}_{\mu}^{\iota}=\sum\limits_{\substack{\mu\text{ is a
partition}\\\text{of }n}}\mathcal{E}_{\mu}^{\iota}=\sum
\limits_{\substack{\lambda\text{ is a partition}\\\text{of }n}%
}\underbrace{\mathcal{E}_{\lambda}^{\iota}}_{=\sum\limits_{\substack{\mu\text{
is a composition of }n\text{;}\\p\left(  \mu\right)  =\lambda}}a_{\mu}^{\iota
}\cdot\iota_{\mu}}\\
&  =\underbrace{\sum\limits_{\substack{\lambda\text{ is a partition}\\\text{of
}n}}\sum\limits_{\substack{\mu\text{ is a composition of }n\text{;}\\p\left(
\mu\right)  =\lambda}}}_{=\sum\limits_{\mu\text{ is a composition of }n}%
}a_{\mu}^{\iota}\cdot\iota_{\mu}=\sum\limits_{\mu\text{ is a composition of
}n}a_{\mu}^{\iota}\cdot\iota_{\mu}=p_{n}%
\end{align*}
). Only if you restrict all maps to the $n$-th graded component of $A$, these
equations become true. Alternatively, you could replace these equations by%
\[
\left(  \sum\limits_{\mu<\left[  n\right]  }\mathcal{E}_{\mu}^{\iota}\right)
^{2}=\left(  p_{n}-\mathcal{E}_{\left[  n\right]  }^{\iota}\right)
^{2}=\underbrace{p_{n}^{2}}_{=p_{n}}-\underbrace{\mathcal{E}_{\left[
n\right]  }^{\iota}\circ p_{n}}_{=\mathcal{E}_{\left[  n\right]  }^{\iota}%
}-\underbrace{p_{n}\circ\mathcal{E}_{\left[  n\right]  }^{\iota}%
}_{=\mathcal{E}_{\left[  n\right]  }^{\iota}}+\underbrace{\left(
\mathcal{E}_{\left[  n\right]  }^{\iota}\right)  ^{2}}_{=\mathcal{E}_{\left[
n\right]  }^{\iota}}=p_{n}-\mathcal{E}_{\left[  n\right]  }^{\iota}%
=\sum\limits_{\mu<\left[  n\right]  }\mathcal{E}_{\mu}^{\iota},
\]
and:%
\[
\left(  \sum\limits_{\mu<\left[  n\right]  }\mathcal{E}_{\mu}^{\iota}\right)
\circ\mathcal{E}_{\left[  n\right]  }^{\iota}=\left(  p_{n}-\mathcal{E}%
_{\left[  n\right]  }^{\iota}\right)  \circ\mathcal{E}_{\left[  n\right]
}^{\iota}=\underbrace{p_{n}\circ\mathcal{E}_{\left[  n\right]  }^{\iota}%
}_{=\mathcal{E}_{\left[  n\right]  }^{\iota}}-\underbrace{\left(
\mathcal{E}_{\left[  n\right]  }^{\iota}\right)  ^{2}}_{=\mathcal{E}_{\left[
n\right]  }^{\iota}}=0.
\]
\newline A similar inaccuracy appears at the end of page 13: There you write
\[
h\circ1=h\circ\left(  h+g+k\right)  =bh+h\circ g.
\]
This is not wrong, but not exactly clear: Probably you want to say%
\[
h=h\circ p_{n}=h\circ\left(  h+g+k\right)  =bh+h\circ g.
\]


\item \textbf{Page 13:} You write: "In other words, $\mathcal{E}_{\left[
n\right]  }^{\iota}$ and $\sum\limits_{\mu<\left[  n\right]  }\mathcal{E}%
_{\mu}^{\iota}$ are two orthogonal idempotents." But in order to show this,
you must not only prove that $\left(  \mathcal{E}_{\left[  n\right]  }^{\iota
}\right)  ^{2}=\mathcal{E}_{\left[  n\right]  }^{\iota}$, $\left(
\sum\limits_{\mu<\left[  n\right]  }\mathcal{E}_{\mu}^{\iota}\right)
^{2}=\sum\limits_{\mu<\left[  n\right]  }\mathcal{E}_{\mu}^{\iota}$ and
$\left(  \sum\limits_{\mu<\left[  n\right]  }\mathcal{E}_{\mu}^{\iota}\right)
\circ\mathcal{E}_{\left[  n\right]  }^{\iota}=0$ (this you have proven), but
also prove that $\mathcal{E}_{\left[  n\right]  }^{\iota}\circ\left(
\sum\limits_{\mu<\left[  n\right]  }\mathcal{E}_{\mu}^{\iota}\right)  =0$.
This is easy, of course:%
\[
\mathcal{E}_{\left[  n\right]  }^{\iota}\circ\left(  \sum\limits_{\mu<\left[
n\right]  }\mathcal{E}_{\mu}^{\iota}\right)  =\mathcal{E}_{\left[  n\right]
}^{\iota}\circ\left(  p_{n}-\mathcal{E}_{\left[  n\right]  }^{\iota}\right)
=\underbrace{\mathcal{E}_{\left[  n\right]  }^{\iota}\circ p_{n}%
}_{=\mathcal{E}_{\left[  n\right]  }^{\iota}}-\underbrace{\left(
\mathcal{E}_{\left[  n\right]  }^{\iota}\right)  ^{2}}_{=\mathcal{E}_{\left[
n\right]  }^{\iota}}=0.
\]
But it should be mentioned, I think.

\item \textbf{Page 14:} You write: "It follows that the coefficients $a_{\mu
}^{\iota^{\epsilon}}$ of the higher Lie idempotents of the third kind depend
polynomially of $\epsilon$."\newline\textit{First,} I don't understand how
this follows from $p_{n}=\sum\limits_{\left\vert \mu\right\vert =n}F_{\mu
}^{\iota^{\epsilon}}$. While all $F_{\mu}^{\iota^{\epsilon}}$ are (by
definition) linear combinations (with constant coefficients) of
\textit{compositions} of various $\iota_{\nu}^{\epsilon}$, it is not clear (to
me) why they are linear combinations (with coefficients polynomial in
$\epsilon$) of \textit{convolutions} of various $\iota_{\nu}^{\epsilon}$. I do
know that $\mathcal{D}_{\iota^{\epsilon}}$ is closed under convolution (by
Theorem 5.1, since $\iota^{\epsilon}\in\left\langle \iota,e\right\rangle
\subseteq\mathcal{D}\left(  A\right)  $), and this yields that they are linear
combinations of convolutions of various $\iota_{\nu}^{\epsilon}$, but why with
coefficients polynomial in $\epsilon$ ?\newline\textit{Second,} even if we can
show that we can write $p_{n}$ as a linear combination of $\iota_{\mu
}^{\epsilon}$ with coefficients polynomial in $\epsilon$, then it is not clear
to me why these coefficients, when specializing at $\epsilon=1$, become our
$a_{\mu}^{\iota}$ - in fact, the $a_{\mu}^{\iota}$ are not always uniquely
determined by $p_{n}=\sum\limits_{\left\vert \mu\right\vert =n}a_{\mu}^{\iota
}\iota_{\mu}$ (since the $\iota_{\mu}$ are not always linearly independent),
so the $a_{\mu}^{\iota}$ you have started with might not be the same as the
$a_{\mu}^{\iota}$ you get by writing $p_{n}$ as a linear combination of
$\iota_{\mu}^{\epsilon}$ and specializing at $\epsilon=1$ (although both
families of $a_{\mu}^{\iota}$ satisfy $p_{n}=\sum\limits_{\left\vert
\mu\right\vert =n}a_{\mu}^{\iota}\iota_{\mu}$).\newline I am interested in how
you actually show that the $a_{\mu}^{\iota^{\epsilon}}$ depend polynomially of
$\epsilon$ in such a way that specialization at $\epsilon=1$ yields our
initial $a_{\mu}^{\iota}$. I think I can show this (with some handwaving)
under the additional condition that $a_{\left[  n\right]  }^{\iota}=1$ for
every $n$. Here is how my proof (roughly) goes:\newline Start with the
equations $p_{n}=\sum\limits_{\left\vert \mu\right\vert =n}a_{\mu}^{\iota
}\iota_{\mu}$. By repeated convolution, these equations yield equations of the
form $p_{\nu}=\sum\limits_{\substack{\left\vert \mu\right\vert =\left\vert
\nu\right\vert ;\\\mu\geq\nu}}a_{\mu,\nu}^{\iota}\iota_{\mu}$ (with
$a_{\mu,\nu}^{\iota}$ being scalars, and $a_{\mu,\left[  n\right]  }^{\iota
}=a_{\mu}^{\iota}$) for all partitions $\nu$, where $\mu\geq\nu$ means that
the composition $\mu$ can be obtained by splitting some parts of $\nu$ into
smaller parts (this defines a partial order $\geq$ on compositions). Since
$a_{\left[  n\right]  }^{\iota}=1$ for every $n$, we find that $a_{\nu,\nu
}^{\iota}=1$ for every composition $\nu$. Now, the equations $p_{\nu}%
=\sum\limits_{\substack{\left\vert \mu\right\vert =\left\vert \nu\right\vert
;\\\mu\geq\nu}}a_{\mu,\nu}^{\iota}\iota_{\mu}$ show us that $\left(
a_{\mu,\nu}^{\iota}\right)  _{\left\vert \mu\right\vert =\left\vert
\nu\right\vert =n}$ is an upper triangular matrix, and the equations
$a_{\nu,\nu}^{\iota}=1$ show that its diagonal entries are $=1$. Hence, it has
an inverse matrix $\left(  b_{\mu,\nu}^{\iota}\right)  _{\left\vert
\mu\right\vert =\left\vert \nu\right\vert =n}$ which satisfies $\iota_{\nu
}=\sum\limits_{\substack{\left\vert \mu\right\vert =\left\vert \nu\right\vert
;\\\mu\geq\nu}}b_{\mu,\nu}^{\iota}p_{\mu}$ for all compositions $\nu$, and
again is upper triangular and has diagonal entries $=1$. The same argument,
done for $e$ instead of $\iota$, shows that there exists a matrix $\left(
b_{\mu,\nu}^{e}\right)  _{\left\vert \mu\right\vert =\left\vert \nu\right\vert
=n}$ which satisfies $e_{\nu}=\sum\limits_{\substack{\left\vert \mu\right\vert
=\left\vert \nu\right\vert ;\\\mu\geq\nu}}b_{\mu,\nu}^{e}p_{\mu}$ for all
compositions $\nu$, and again is upper triangular and has its diagonal entries
$=1$. Now, the matrix $\left(  \epsilon\cdot b_{\mu,\nu}^{\iota}+\left(
1-\epsilon\right)  \cdot b_{\mu,\nu}^{e}\right)  _{\left\vert \mu\right\vert
=\left\vert \nu\right\vert =n}$ satisfies%
\begin{align*}
\iota_{\nu}^{\epsilon} &  =\epsilon\cdot\iota_{\nu}+\left(  1-\epsilon\right)
\cdot e_{\nu}=\epsilon\cdot\sum\limits_{\substack{\left\vert \mu\right\vert
=\left\vert \nu\right\vert ;\\\mu\geq\nu}}b_{\mu,\nu}^{\iota}p_{\mu}+\left(
1-\epsilon\right)  \cdot\sum\limits_{\substack{\left\vert \mu\right\vert
=\left\vert \nu\right\vert ;\\\mu\geq\nu}}b_{\mu,\nu}^{e}p_{\mu}\\
&  =\sum\limits_{\substack{\left\vert \mu\right\vert =\left\vert
\nu\right\vert ;\\\mu\geq\nu}}\left(  \epsilon\cdot b_{\mu,\nu}^{\iota
}+\left(  1-\epsilon\right)  \cdot b_{\mu,\nu}^{e}\right)  p_{\mu}%
\end{align*}
for all compositions $\nu$, and again is upper triangular and has its diagonal
entries $=1$. Hence, its inverse matrix $\left(  a_{\mu,\nu}^{\iota^{\epsilon
}}\right)  _{\left\vert \mu\right\vert =\left\vert \nu\right\vert =n}$
satisfies $p_{n}=\sum\limits_{\left\vert \mu\right\vert =n}a_{\mu,\left[
n\right]  }^{\iota^{\epsilon}}\iota_{\mu}^{\epsilon}$, but its entries
$a_{\mu,\nu}^{\iota^{\epsilon}}$ are polynomials in the entries of $\left(
\epsilon\cdot b_{\mu,\nu}^{\iota}+\left(  1-\epsilon\right)  \cdot b_{\mu,\nu
}^{e}\right)  _{\left\vert \mu\right\vert =\left\vert \nu\right\vert =n}$
(because if $C$ is an upper triangular matrix with diagonal entries $=1$, then
the entries of $C^{-1}$ are polynomials in the entries of $C$), and thus
polynomials in $\epsilon$. This gives us what we want.\newline But I cannot
get rid of the condition that $a_{\left[  n\right]  }^{\iota}=1$ for every $n$
(not only for the one we are working with, but also for the smaller $n$,
because we need all $a_{\nu,\nu}^{\iota}$ to be $1$).\newline\textbf{HOWEVER},
I think that I can modify your proof of Theorem 5.4 in a different way to make
it valid:\newline First of all, let us generalize the results of Section 3
from one Lie idempotent to two Lie idempotents:\footnote{In the following
Lemmas 5.6, 5.7, 5.8 and 5.9, we don't assume that $\mathcal{D}\left(
A\right)  =\mathcal{D}_{\iota}$.}\newline\textbf{Lemma 5.6.} Let $\iota$ and
$\rho$ be two Lie idempotents. Then, any two compositions $\lambda$ and $\mu$
such that $\left\vert \lambda\right\vert \neq\left\vert \mu\right\vert $
satisfy $\iota_{\lambda}\circ\rho_{\mu}=0$.\newline This is a very obvious
fact (it is obvious because the image of $\rho_{\mu}$ lies in the $\left\vert
\mu\right\vert $-th graded component of $H$, whereas $\iota_{\lambda}$ sends
every graded component of $H$ except of the $\left\vert \lambda\right\vert
$-th one to $0$), and it generalizes the property $\iota_{\lambda}\circ
\iota_{\mu}=0$ for $\left\vert \lambda\right\vert \neq\left\vert
\mu\right\vert $.\newline Less trivially, we have:\newline\textbf{Lemma 5.7.}
Let $\iota$ and $\rho$ be two Lie idempotents. Let $\mu$ and $\lambda$ be two
compositions of the same weight and the same length $k$.\newline\textbf{(i)}
If $p\left(  \lambda\right)  \neq p\left(  \mu\right)  $, then $\iota_{\mu
}\circ\rho_{\lambda}=0$.\newline\textbf{(ii)} If $p\left(  \lambda\right)
=p\left(  \mu\right)  $, then $\iota_{\mu}\circ\rho_{\lambda}=N\rho_{\mu}$,
where $N$ is the number of permutations of $\left\{  1,2,...,k\right\}  $
which act trivially on the sequence $p\left(  \mu\right)  =p\left(
\lambda\right)  $. (This number $N$ only depends on $p\left(  \lambda\right)
=p\left(  \mu\right)  $, and will often be denoted by $N\left(  p\left(
\lambda\right)  \right)  $ or by $N\left(  \lambda\right)  $.)\newline For the
proof of Lemma 5.7, proceed in the same way as in the proof of Lemma 3.1. You
will need the identity $\iota\circ\rho=\rho$, which follows from $\iota
\mid_{\operatorname*{Prim}A}=\operatorname*{id}\nolimits_{\operatorname*{Prim}%
A}$ (because both $\iota$ and $\rho$ are Lie idempotents, i. e., projections
on $\operatorname*{Prim}A$).\newline Similarly:\newline\textbf{Lemma 5.8.} Let
$\iota$ and $\rho$ be two Lie idempotents. Let $\mu$ and $\lambda$ be two
compositions of the same weight such that $l\left(  \mu\right)  >l\left(
\lambda\right)  $. Then $\iota_{\mu}\circ\rho_{\lambda}=0$.\newline This is
proven in the same way as Lemma 3.2.\newline Next, we need a kind of
generalization of Lemma 5.3:\newline\textbf{Lemma 5.9.} Let $\iota$ and $\rho$
be two Lie idempotents. Let $\lambda$ be a partition. For every composition
$\mu$ with $p\left(  \mu\right)  =\lambda$, let $b_{\mu}^{\iota}$ and $b_{\mu
}^{\rho}$ be two scalars. Then,%
\[
\left(  \sum\limits_{p\left(  \mu\right)  =\lambda}b_{\mu}^{\iota}\iota_{\mu
}\right)  \circ\left(  \sum\limits_{p\left(  \mu\right)  =\lambda}b_{\mu
}^{\rho}\rho_{\mu}\right)  =\left(  \sum\limits_{p\left(  \mu\right)
=\lambda}b_{\mu}^{\iota}\right)  N\left(  \sum\limits_{p\left(  \mu\right)
=\lambda}b_{\mu}^{\rho}\rho_{\mu}\right)  ,
\]
where $N$ is the number of permutations of $\left\{  1,2,...,k\right\}  $
which act trivially on the sequence $\lambda$.\newline The proof of this lemma
proceeds in the same way as the identity $\left(  \sum\limits_{p\left(
\mu\right)  =\lambda}b_{\mu}\iota_{\mu}\right)  ^{2}=\left(  \sum
\limits_{p\left(  \mu\right)  =\lambda}b_{\mu}\right)  N\left(  \sum
\limits_{p\left(  \mu\right)  =\lambda}b_{\mu}\iota_{\mu}\right)  $ was proven
in the proof of Lemma 5.3. Here are the details of the proof:

\textit{Proof of Lemma 5.9.} For every composition $\mu$ satisfying $p\left(
\mu\right)  =\lambda$, we know that $N$ is the number of permutations of
$\left\{  1,2,...,k\right\}  $ which act trivially on the sequence $p\left(
\mu\right)  $ (because $N$ is defined as the number of permutations of
$\left\{  1,2,...,k\right\}  $ which act trivially on the sequence $\lambda$,
but we have $\lambda=p\left(  \mu\right)  $). Hence, for every composition
$\mu$ satisfying $p\left(  \mu\right)  =\lambda$, we have $i_{\mu}\circ
\rho_{\mu}=N\rho_{\mu}$ (by Lemma 5.7 \textbf{(ii)}, applied to $\mu$ instead
of $\lambda$). Since composition of linear maps is bilinear, we have%
\begin{align*}
&  \left(  \sum\limits_{p\left(  \mu\right)  =\lambda}b_{\mu}^{\iota}%
\iota_{\mu}\right)  \circ\left(  \sum\limits_{p\left(  \mu\right)  =\lambda
}b_{\mu}^{\rho}\rho_{\mu}\right)  \\
&  =\sum\limits_{p\left(  \mu\right)  =\lambda}\sum\limits_{p\left(
\mu\right)  =\lambda}b_{\mu}^{\iota}b_{\mu}^{\rho}\underbrace{\iota_{\mu}%
\circ\rho_{\mu}}_{=N\rho_{\mu}}=N\sum\limits_{p\left(  \mu\right)  =\lambda
}\sum\limits_{p\left(  \mu\right)  =\lambda}b_{\mu}^{\iota}b_{\mu}^{\rho}%
\rho_{\mu}\\
&  =\left(  \sum\limits_{p\left(  \mu\right)  =\lambda}b_{\mu}^{\iota}\right)
N\left(  \sum\limits_{p\left(  \mu\right)  =\lambda}b_{\mu}^{\rho}\rho_{\mu
}\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{since composition of linear maps
is bilinear}\right)  .
\end{align*}
This proves Lemma 5.9.\newline Now to the \textit{proof of Theorem 5.4.} We
proceed in the same way as you do (with one exception: we don't have to assume
$h\neq0$) until your Claim 5.5 (which we cannot make anymore, since we haven't
assumed that $h\neq0$). Then, just as you, we prove $h\circ g=\left(
1-b\right)  h$ and $k\circ g=\left(  b-1\right)  h$. Now I am going to show
that $h^{2}=h$.\newline First of all, we have $p_{n}=\sum\limits_{\left\vert
\mu\right\vert =n}\dfrac{1}{n!}e_{\mu}$\ \ \ \ \footnote{This is a known fact
(I knew it in the form $p_{n}=\sum\limits_{\ell=0}^{n}\dfrac{1}{\ell!}%
\sum\limits_{\substack{\left(  a_{1},a_{2},...,a_{\ell}\right)  \in\left\{
1,2,...,n\right\}  ^{\ell};\\n=a_{1}+a_{2}+...+a_{\ell}}}\left(  e_{a_{1}}\ast
e_{a_{2}}\ast...\ast e_{a_{\ell}}\right)  $). It can be easily derived from
the fact that $e=\log_{\ast}\left(  \operatorname*{id}\right)  $, so that
$\operatorname*{id}=\exp_{\ast}e=\exp_{\ast}\left(  e_{1}+e_{2}+e_{3}%
+...\right)  $.}. Let us define a scalar $a_{\mu}^{e}$ by $a_{\mu}^{e}%
=\dfrac{1}{n!}$ for every partition $\mu$. Then, $p_{n}=\sum
\limits_{\left\vert \mu\right\vert =n}\underbrace{\dfrac{1}{n!}}_{=a_{\mu}%
^{e}}e_{\mu}=\sum\limits_{\left\vert \mu\right\vert =n}a_{\mu}^{e}e_{\mu}$.
Hence, in the same way as we defined an element $\mathcal{E}_{\lambda}^{\iota
}$ for every partition $\lambda$ in Definition 5.2, we can define an element
$\mathcal{E}_{\lambda}^{e}$ for every partition $\lambda$ by the formula%
\[
\mathcal{E}_{\lambda}^{e}:=\sum\limits_{p\left(  \mu\right)  =\lambda
}\underbrace{a_{\mu}^{e}}_{=\dfrac{1}{n!}}\cdot e_{\mu}=\sum\limits_{p\left(
\mu\right)  =\lambda}\dfrac{1}{n!}e_{\mu}.
\]
\newline From Lemmas 5.7 and 5.8 (applied to $e$ and $\iota$ instead of
$\iota$ and $\rho$), we conclude that $\mathcal{E}_{\lambda}^{e}%
\circ\mathcal{E}_{\mu}^{\iota}=0$ for every partition $\mu<\lambda$. Hence,%
\[
\mathcal{E}_{\lambda}^{e}\circ\underbrace{k}_{=\sum\limits_{\mu<\lambda
}\mathcal{E}_{\mu}^{\iota}}=\mathcal{E}_{\lambda}^{e}\circ\left(
\sum\limits_{\mu<\lambda}\mathcal{E}_{\mu}^{\iota}\right)  =\sum
\limits_{\mu<\lambda}\underbrace{\mathcal{E}_{\lambda}^{e}\circ\mathcal{E}%
_{\mu}^{\iota}}_{\substack{=0\\\text{(since }\mu<\lambda\text{)}}}=0.
\]
On the other hand, for every partition $\lambda$, let $N\left(  \lambda
\right)  $ denote the number of permutations of $\left\{  1,2,...,k\right\}  $
which act trivially on the sequence $\lambda$. We have $\mathcal{E}_{\lambda
}^{e}=\sum\limits_{p\left(  \mu\right)  =\lambda}\dfrac{1}{n!}e_{\mu}$ and
$h=\mathcal{E}_{\lambda}^{\iota}=\sum\limits_{p\left(  \mu\right)  =\lambda
}a_{\mu}^{\iota}e_{\mu}$, so that%
\begin{align*}
\mathcal{E}_{\lambda}^{e}\circ h &  =\left(  \sum\limits_{p\left(  \mu\right)
=\lambda}\dfrac{1}{n!}e_{\mu}\right)  \circ\left(  \sum\limits_{p\left(
\mu\right)  =\lambda}a_{\mu}^{\iota}\iota_{\mu}\right)  =\left(
\sum\limits_{p\left(  \mu\right)  =\lambda}\dfrac{1}{n!}\right)  N\left(
\lambda\right)  \cdot\underbrace{\left(  \sum\limits_{p\left(  \mu\right)
=\lambda}a_{\mu}^{\iota}\iota_{\mu}\right)  }_{=h}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by Lemma 5.9, applied to }N\left(
\lambda\right)  \text{, }\dfrac{1}{n!}\text{, }a_{\mu}^{\iota}\text{, }e\text{
and }\iota\text{ instead of }N\text{, }b_{\mu}^{\iota}\text{, }b_{\mu}^{\rho
}\text{, }\iota\text{ and }\rho\right)  \\
&  =\left(  \sum\limits_{p\left(  \mu\right)  =\lambda}\dfrac{1}{n!}\right)
N\left(  \lambda\right)  h.
\end{align*}
Now, compare%
\[
\underbrace{\mathcal{E}_{\lambda}^{e}\circ k}_{=0}\circ g=0\circ g=0
\]
with%
\[
\mathcal{E}_{\lambda}^{e}\circ\underbrace{k\circ g}_{=\left(  b-1\right)
h}=\left(  b-1\right)  \underbrace{\mathcal{E}_{\lambda}^{e}\circ h}_{=\left(
\sum\limits_{p\left(  \mu\right)  =\lambda}\dfrac{1}{n!}\right)  N\left(
\lambda\right)  h}=\left(  b-1\right)  \left(  \sum\limits_{p\left(
\mu\right)  =\lambda}\dfrac{1}{n!}\right)  N\left(  \lambda\right)  h.
\]
This yields%
\[
\left(  b-1\right)  \left(  \sum\limits_{p\left(  \mu\right)  =\lambda}%
\dfrac{1}{n!}\right)  N\left(  \lambda\right)  h=0.
\]
Since $\left(  \sum\limits_{p\left(  \mu\right)  =\lambda}\dfrac{1}%
{n!}\right)  N\left(  \lambda\right)  $ is invertible in $k$ (in fact,
$\left(  \sum\limits_{p\left(  \mu\right)  =\lambda}\dfrac{1}{n!}\right)
N\left(  \lambda\right)  \neq0$ obviously; we can even prove that $\left(
\sum\limits_{p\left(  \mu\right)  =\lambda}\dfrac{1}{n!}\right)  N\left(
\lambda\right)  =1$, but we don't need this), this becomes $\left(
b-1\right)  h=0$, so that $h=bh$. Compared with $h\circ h=bh$ (which follows
from the proof of Lemma 5.3), this yields $h\circ h=h$, so that $h$ is an
idempotent.\newline Since $g^{2}=g$ (because $g=\sum\limits_{\mu>\lambda
}\mathcal{E}_{\mu}^{\iota}$, and by the induction assumption the
$\mathcal{E}_{\mu}^{\iota}$ are orthogonal idempotents), $h\circ g=\left(
1-b\right)  h=-\underbrace{\left(  b-1\right)  h}_{=0}=0$ and $k\circ
g=\left(  b-1\right)  h=0$, we can continue the proof as you do after you
prove Claim 5.5. This proves Theorem 5.4.

\item \textbf{Page 14:} There is a typo: $b_{\lambda}^{i^{\epsilon}}$ should
be $b_{\lambda}^{\iota^{\epsilon}}$.

\item \textbf{Page 15:} You write: "and the proof of theorem 5.3 is complete".
The theorem is Theorem 5.4, not 5.3.

\item \textbf{Page 16:} In reference [R1], typo: "represntations".
\end{itemize}


\end{document}