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\ihead{Errata to ``From iterated integrals...''}
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\begin{document}

\begin{center}
\textbf{From iterated integrals and chronological calculus to Hopf and
Rota-Baxter algebras}

\textit{Kurusch Ebrahimi-Fard, Fr\'{e}d\'{e}ric Patras}

\href{https://arxiv.org/abs/1911.08766v1}{arXiv:1911.08766v1}

\textbf{Errata and addenda by Darij Grinberg}

\bigskip
\end{center}

\begin{enumerate}
\item \textbf{page 5, very last display:} The $\int\limits_{0\leq t_{1}\leq
t_{2}\leq t}$ sign should be a double integral sign: $\iint\limits_{0\leq
t_{1}\leq t_{2}\leq t}$.

\item \textbf{page 6, middle of the page (two paragraphs above Definition 2):}
What exactly do you mean by \textquotedblleft(looking for example at the
boundary of $\Delta_{t}^{\nu}$)\textquotedblright?

\item \textbf{page 7, first display:} Here I would also add the alternative
definition%
\[
\mathbf{L}_{\sigma}\left(  t\right)  =\iiint\limits_{0\leq t_{1}\leq t_{2}%
\leq\cdots\leq t_{n}\leq t}dt_{1}\cdots dt_{n}L_{1}\left(  t_{\sigma\left(
1\right)  }\right)  \cdots L_{n}\left(  t_{\sigma\left(  n\right)  }\right)
.
\]
This is the form you often use below.

\item \textbf{page 7, Lemma 3:} This looks wrong. Shouldn't it be
$\mathbf{L}_{\alpha}\left(  t\right)  \mathbf{K}_{\beta}\left(  t\right)
=\left(  \mathbf{L}\cdot\mathbf{K}\right)  _{\alpha\ast\beta}\left(  t\right)
$ ?

\item \textbf{page 7, Lemma 4:} Replace \textquotedblleft$\gamma\left(
n+1\right)  =n+1$\textquotedblright\ by \textquotedblleft$\gamma\left(
n+1\right)  =n+m+1$\textquotedblright, or so I believe.

\item \textbf{page 8, Proposition 8:} \textquotedblleft
basis\textquotedblright\ $\rightarrow$ \textquotedblleft
bases\textquotedblright.

\item \textbf{page 8, Remark 10:} \textquotedblleft availble\textquotedblright.

\item \textbf{page 8, Theorem 11:} Here, you are letting the group algebra of
the symmetric group act \textquotedblleft formally\textquotedblright\ on the
operators, to make sense of expressions like $X_{n}\left(  t\right)  \cdot w$
for some $w$ in the group algebra. This is worth explaining.

\item \textbf{page 9, proof of Theorem 11:} Shouldn't the \textquotedblleft%
$\left(  -1\right)  ^{l-1}$\textquotedblright\ be a \textquotedblleft$\left(
-1\right)  ^{l}$\textquotedblright?

\item \textbf{page 9, proof of Theorem 11:} The last equality sign here
behooves more explanation, I believe.

\item \textbf{page 9, Remark 12:} Shouldn't the \textquotedblleft%
$321$\textquotedblright\ be \textquotedblleft$312$\textquotedblright?

\item \textbf{page 10:} \textquotedblleft branche\textquotedblright%
\ $\rightarrow$ \textquotedblleft branch\textquotedblright.

\item \textbf{page 10:} In \textquotedblleft inequalities $t_{i}\leq
t_{j}\Longleftrightarrow j\leq_{T}i$\textquotedblright, the \textquotedblleft%
$\Longleftrightarrow$\textquotedblright\ sign should be a \textquotedblleft%
$\Longleftarrow$\textquotedblright\ sign, right?

\item \textbf{page 12, last paragraph:} The connection between Chen's rule and
(6) is not very clear to me.

\item \textbf{page 13, Corollary 24:} \textquotedblleft over all
admissible\textquotedblright\ $\rightarrow$ \textquotedblleft over all
nonempty admissible\textquotedblright.

\item \textbf{page 14, first displayed equation:} The right hand side has a
closing parenthesis $)$ too much.

\item \textbf{page 16, Definition 25:} To me, the word \textquotedblleft
shuffle algebra\textquotedblright\ means the bialgebra $V^{\otimes0}\oplus
V^{\otimes1}\oplus V^{\otimes2}\oplus\cdots$ with the shuffle product as
multiplication and with deconcatenation as comultiplication. What you call a
\textquotedblleft shuffle algebra\textquotedblright\ is what I know as a
\textquotedblleft dendriform algebra\textquotedblright. Wouldn't it be good to
point out the alternative language here rather than just in Remark 30?

\item \textbf{page 16, Remark 27:} Maybe better to use the $\prec$ and $\succ$
symbols for the operations here, seeing that this remark is about general
shuffle algebras rather than just the integral ones.

\item \textbf{page 17, proof sketch to Lemma 29:} In the second displayed
equation, the right hand side should be $\left(  \left(  y\succ x\prec
z\right)  \cdot_{A}y^{\prime}\right)  \succ x\prec z^{\prime}$. (So no two
$x$'es, and no extra $)$ parenthesis at the end.)

\item \textbf{page 17, after Definition 31:} \textquotedblleft
Caley\textquotedblright\ $\rightarrow$ \textquotedblleft
Cayley\textquotedblright.

\item \textbf{page 19, Definition 35:} Add \textquotedblleft$\left\{
{}\right\}  v=v$\textquotedblright, since otherwise the brace map is not
defined on the $0$-th degree component of $T\left(  V\right)  $.

\item \textbf{page 19, Definition 35:} Does \textquotedblleft$\left\{
w_{1},\ldots,w_{n}\right\}  $\textquotedblright\ really mean \textquotedblleft%
$\left\{  w_{1}\ldots w_{n}\right\}  $\textquotedblright?

I also don't see an immediate reason why this is well-defined. Symmetric
tensors like $w_{1}\ldots w_{n}$ come with no canonical ordering of their
factors, so $w_{n}$ could as well be any other $w_{i}$, but the right hand
side is not obviously symmetric. I see a long proof by induction on $n$ (using
the pre-Lie axiom), but this should be cited to a place in the literature.

\item \textbf{page 19:} \textquotedblleft Recall that the time-ordered
product\textquotedblright: Why \textquotedblleft Recall\textquotedblright? Has
the time-ordered product been defined before?

\item \textbf{page 20, Theorem 38:} Shouldn't \textquotedblleft%
$V:=\mathcal{PL}\left(  L\right)  $\textquotedblright\ be \textquotedblleft%
$L:=\mathcal{PL}\left(  V\right)  $\textquotedblright?

\item \textbf{page 21, second paragraph:} In the definition of $G\left(
\widehat{H}\right)  $, add the extra condition $\epsilon\left(  x\right)  =1$
(otherwise, $0$ would be an element).

\item \textbf{page 21, second paragraph:} \textquotedblleft Notice that
$\widehat{k\left[  V\right]  }$ is such a Hopf algebra\textquotedblright%
\ $\rightarrow$ \textquotedblleft Notice that $\widehat{k\left[  V\right]  }$
is such a completed Hopf algebra\textquotedblright.

\item \textbf{pages 20--21:} Do you really need $V$ to be a \textbf{free}
pre-Lie algebra for these Magnus computations to work? I don't understand them
in detail, but my impression is that any graded (in positive degrees) pre-Lie
algebra would work.

\item \textbf{page 21, Definition 41:} You don't use the words
\textquotedblleft$2$-group\textquotedblright\ and \textquotedblleft$2$-pre-Lie
algebra\textquotedblright\ anywhere outside of this definition. (I also would
suggest using something other than \textquotedblleft$2$%
-group\textquotedblright, as that word already has two different meanings --
sadly, $2$ is prime...)

\item \textbf{page 21, Definition 41:} Shouldn't the $\vartriangleleft$ sign
be a $\vartriangleright$ sign?

\item \textbf{page 26, Lemma 47:} This requires the algebra to be commutative,
or at least some weaker version thereof. For example, for $n=1$ and $m=1$,
this lemma is saying that $R\left(  f\right)  R\left(  g\right)  =R\left(
fR\left(  g\right)  \right)  +R\left(  gR\left(  f\right)  \right)  $, which
differs from the Rota--Baxter identity (17) in that $R\left(  R\left(
f\right)  g\right)  $ is replaced by $R\left(  gR\left(  f\right)  \right)  $.

\item \textbf{page 27, Example 50:} \textquotedblleft In fact, any projector
which satisfies relation (15) is of weight $\theta=1$\textquotedblright: This
is not literally true, since (e.g.) a projector onto a square-zero ideal along
a subalgebra will satisfy the relation (15) for any $\theta$.

\item \textbf{page 27, last paragraph:} I don't see why you write
\textquotedblleft$\left(  m,n;r\right)  $-quasi-shuffle of type $\max\left(
m,n\right)  \leq r\leq m+n$\textquotedblright\ when all the information is
contained in the \textquotedblleft$\left(  m,n;r\right)  $%
-quasi-shuffle\textquotedblright\ part. In this context, the \textquotedblleft
type\textquotedblright\ part is only confusing. Also, the first time you use
the formulation, you misspell the second \textquotedblleft$\leq$%
\textquotedblright\ sign as a \textquotedblleft$<$\textquotedblright\ sign,
which is wrong.

\item \textbf{page 28, Lemma 52:} The \textquotedblleft$x_{i}$%
\textquotedblright\ at the end should be an \textquotedblleft$x_{j}%
$\textquotedblright.

\item \textbf{page 28, Definition 53:} \textquotedblleft operator on
$A$\textquotedblright\ $\rightarrow$ \textquotedblleft operator $R$ on
$A$\textquotedblright.

\item \textbf{page 30, Lemma 64:} The comma after \textquotedblleft weight
$\theta$\textquotedblright\ should be a period.

\item \textbf{page 31:} \textquotedblleft possesses more
properties\textquotedblright\ $\rightarrow$ \textquotedblleft possess more
properties\textquotedblright.

\item \textbf{page 31:} \textquotedblleft seen in Theorem 61\textquotedblright%
\ $\rightarrow$ \textquotedblleft seen in Proposition 61\textquotedblright%
\ (or turn the proposition into a theorem).

\item \textbf{page 32, (37):} What are these equations for? Are the two
\textquotedblleft exp\textquotedblright\ elements sought? Are they supposed to
live in $A\left[  \left[  \lambda\right]  \right]  $ ?

\item \textbf{page 34, (39):} The sum on the left hand side should be over
\textquotedblleft$m>0$\textquotedblright, not over \textquotedblleft%
$n>0$\textquotedblright.

\item \textbf{page 35, after Proposition 73:} Why is the \textquotedblleft
obvious identity\textquotedblright\ obvious?

\item \textbf{page 36, (45):} I assume the $\prod_{i=1}^{m_{j}-1}$ product is
taken in the order of $i$ increasing from left to right?

\item \textbf{page 36, \S 4.6:} With just a bit more work, you can prove many
of the claims stated here.

To wit, consider the free commutative Rota--Baxter algebra $RB\left(
x\right)  $ on a single generator $x$. Let $\Psi:RB\left(  x\right)
\rightarrow A^{\mathbb{N}}$ be the Rota--Baxter algebra homomorphism sending
$x$ to $x$ (of course, a different $x$). Let the \emph{chain monomials} be the
elements of the form $x^{a_{0}}R\left(  x^{a_{1}}R\left(  x^{a_{2}}R\left(
\cdots R\left(  x^{a_{k}}\right)  \right)  \right)  \right)  $ in $RB\left(
x\right)  $ for $\left(  a_{0},a_{1},\ldots,a_{k}\right)  \in\mathbb{N}^{k}$
(where $0\in\mathbb{N}$). It is easy to see (by induction on $k$) that the
image of such a standard monomial $x^{a_{0}}R\left(  x^{a_{1}}R\left(
x^{a_{2}}R\left(  \cdots R\left(  x^{a_{k}}\right)  \right)  \right)  \right)
$ under $\Psi$ is the sequence $\left(  u_{0},u_{1},u_{2},\ldots\right)  \in
A^{\mathbb{N}}$, where%
\[
u_{i}=\sum_{i=j_{0}>j_{1}>j_{2}>\cdots>j_{k}}x_{j_{0}}^{a_{0}}x_{j_{1}}%
^{a_{1}}\cdots x_{j_{k}}^{a_{k}}\ \ \ \ \ \ \ \ \ \ \text{for each }%
i\in\mathbb{N}.
\]
These images are linearly independent (since you can reconstruct $\left(
a_{0},a_{1},\ldots,a_{k}\right)  $ from the monomials appearing in $u_{k+1}$).
Thus, the chain monomials in $RB\left(  x\right)  $ are also linearly
independent. Since they furthermore span $RB\left(  x\right)  $ (since Lemma
52 lets us expand the product of two such monomials as a sum of such
monomials, and it is easy to apply $R$ to such a monomial), it follows that
they form a basis of $RB\left(  x\right)  $. Since their images under $\Psi$
are linearly independent, this entails that the map $\Psi$ is injective, and
therefore the Rota--Baxter subalgebra of $A^{\mathbb{N}}$ generated by $x$
(that is, the image of $\Psi$) is also free as a commutative Rota--Baxter algebra.

Let $\Gamma:\mathcal{S}\rightarrow RB\left(  x\right)  $ be the algebra
homomorphism that sends each $e_{n}\in\mathcal{S}$ to $R^{\left(  n\right)
}\left(  x\right)  \in RB\left(  x\right)  $. This is the embedding of
$\mathcal{S}$ into $RB\left(  x\right)  $ that you are talking about, but we
have yet to prove that it is an embedding.

Now, let $\Phi:\mathcal{S}\rightarrow A^{\mathbb{N}}$ be the map that sends
each $f\in\mathcal{S}$ to the sequence $\left(  v_{0},v_{1},v_{2}%
,\ldots\right)  \in A^{\mathbb{N}}$, where%
\[
v_{i}=f\left(  x_{0},x_{1},\ldots,x_{i-1},0,0,0,\ldots\right)
\ \ \ \ \ \ \ \ \ \ \text{for each }i\in\mathbb{N}.
\]
This is also an algebra homomorphism, and is easily seen to be injective.
Moreover, it is easy to see (by induction on $n$) that $\Phi\left(
e_{n}\right)  =R^{\left(  n\right)  }\left(  x\right)  $ for each
$n\in\mathbb{N}$. Hence, it follows easily that $\Phi=\Psi\circ\Gamma$. This
shows that $\Gamma$ is injective (since $\Phi$ is injective), i.e., an embedding.

It is moreover trivial to see that $\Phi\left(  h_{n}\right)  =R\left(
x^{n}\right)  $ for each $n>0$ for the power-sum symmetric functions $h_{n}$
(which everyone I know calls $p_{n}$, by the way). Therefore, by
\textquotedblleft un-applying\textquotedblright\ the injective map $\Psi$, we
obtain $\Gamma\left(  h_{n}\right)  =R\left(  x^{n}\right)  $ as well (since
$\Psi$ sends $\Gamma\left(  h_{n}\right)  $ to $\Phi\left(  h_{n}\right)  $
and sends $R\left(  x^{n}\right)  \in RB\left(  x\right)  $ to $R\left(
x^{n}\right)  \in A^{\mathbb{N}}$). This is one of the properties of the
embedding/correspondence between $\mathcal{S}$ and $RB\left(  x\right)  $ that
you have stated.

(I have learned the above ideas in some form from two papers by Glanffrwd
Thomas: \url{https://doi.org/10.1007/bfb0090016} and
\url{https://doi.org/10.1016/B978-0-12-428780-8.50012-X} .)

\item \textbf{page 37, Definition 74:} \textquotedblleft by the $R^{\left(
n\right)  }\left(  x\right)  $\textquotedblright\ should be \textquotedblleft
by the $R\left(  x^{n}\right)  $\textquotedblright\ here.

\item \textbf{page 40, Definition 77:} \textquotedblleft by the $R^{\left(
n\right)  }\left(  x\right)  $\textquotedblright\ should be \textquotedblleft
by the $R\left(  x^{n}\right)  $\textquotedblright\ here in the second paragraph.

However, there is another problem: I don't see how to express $R\left(
x^{2}\right)  $ in terms of the generators $R^{\left(  n\right)  }\left(
x\right)  $, since the equality $\underbrace{R\left(  x\right)  \cdot R\left(
x\right)  }_{=\left(  R^{\left(  1\right)  }\left(  x\right)  \right)  ^{2}%
}=\underbrace{R\left(  R\left(  x\right)  x\right)  }_{=R^{\left(  2\right)
}\left(  x\right)  }+R\left(  xR\left(  x\right)  \right)  +R\left(
x^{2}\right)  $ contains two \textquotedblleft unknowns\textquotedblright\ (we
can no longer equate $R\left(  xR\left(  x\right)  \right)  $ with $R\left(
R\left(  x\right)  x\right)  $). What am I missing?

\item \textbf{References:} Why is the title of a paper sometimes formatted in
\textit{italics} and sometimes in \textsl{slanted}?
\end{enumerate}


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