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\ihead{Errata to ``Schubert polynomial expansions revisited''}
\ohead{\today}
\begin{document}

\begin{center}
\textbf{Schubert polynomial expansions revisited}

\textit{Philippe Nadeau, Hunter Spink and Vasu Tewari}

\href{https://doi.org/10.1017/fms.2025.10068}{doi:10.1017/fms.2025.10068}

\textbf{Errata and addenda by Darij Grinberg}

\bigskip
\end{center}

I will refer to the results appearing in the article \textquotedblleft
Schubert polynomial expansions revisited\textquotedblright\ by the numbers
under which they appear in this article (specifically, in its
\href{https://doi.org/10.1017/fms.2025.10068}{published version
(doi:10.1017/fms.2025.10068)}).

\setcounter{section}{7}

\section{Errata}

\begin{enumerate}
\item \textbf{Page 3:} The word \textquotedblleft
immediately\textquotedblright\ one line above Theorem 1.1 appears to be
somewhat of an overstatement...

\item \textbf{Page 6, \S 2.1:} It should be added that the value $\ell\left(
w\right)  $ of the length function $\ell$ at a given element $w\in M$ is
called the \textit{length} or the \textit{degree} of $w$ (the two words are
being used rather synonymously here).

\item \textbf{Page 6, \S 2.1:} I think you want to require $M$ to be
\textbf{equipped with} a specific enumeration $\left(  a_{1},a_{2}%
,a_{3},\ldots\right)  $ of $M_{1}$ (that is, a countably infinite list
$\left(  a_{1},a_{2},a_{3},\ldots\right)  $ of distinct elements of $M_{1}$
that contains each of these elements exactly once). This way, when you refer
to $\left\{  a_{i}\right\}  _{i\geq1}$ in Definition 2.4, it is clear what the
$a_{i}$ are, and the condition $\max\operatorname*{supp}c\left(  w\right)
=\max\operatorname*{Last}\left(  w\right)  $ on page 9 makes sense (since
$\operatorname*{Last}\left(  w\right)  $ is a set of positive integers just
like $\operatorname*{supp}c\left(  w\right)  $).

\item \textbf{From page 6,\ \S 2.1 onwards:} The word \textquotedblleft
right-cancellative\textquotedblright\ is sometimes spelled \textquotedblleft
right-cancellable\textquotedblright\ (once on page 4 and once on page 14). I
think the first spelling is more reasonable, but they are both fine, as long
as the synonymity is mentioned.

\item \textbf{Page 6, \S 2.2:} It should be said that $\operatorname*{Pol}%
\nolimits_{n}$ denotes the polynomial ring $\mathbb{Z}\left[  x_{1}%
,x_{2},\ldots,x_{n}\right]  $.

\item \textbf{Page 7, Example 2.7:} Am I confused or have you never properly
defined the monoids $\mathsf{Codes}$ ? As I guess, it should be defined as the
monoid of all finitely supported infinite sequences $\left(  c_{i}\right)
_{i\geq1}=\left(  c_{1},c_{2},c_{3},\ldots\right)  $ of nonnegative integers
(where \textquotedblleft finitely supported\textquotedblright\ means that all
but finitely many entries are $0$). Its multiplication is entrywise addition
of sequences: i.e.,%
\[
\left(  c_{i}\right)  _{i\geq1}\left(  d_{i}\right)  _{i\geq1}:=\left(
c_{i}+d_{i}\right)  _{i\geq1}\ \ \ \ \ \ \ \ \ \ \text{for any }\left(
c_{i}\right)  _{i\geq1},\left(  d_{i}\right)  _{i\geq1}\in\mathsf{Codes}.
\]


As an abelian monoid, it is freely generated by the \textquotedblleft standard
basis vectors\textquotedblright\ $e_{i}:=\left(  0,0,\ldots,0,1,0,0,0,\ldots
\right)  $ (with the $1$ in the $i$-th position) for all $i\geq1$. By abuse of
notation, we denote these basis vectors $e_{i}$ simply by $i$ in Example 2.7,
so that $\mathsf{Codes}$ is the free abelian monoid with free generators
$1,2,3,\ldots$.

\item \textbf{Page 7, Example 2.10:} \textquotedblleft to to\textquotedblright%
\ $\rightarrow$ \textquotedblleft to\textquotedblright.

\item \textbf{Page 8, proof of Proposition 2.11:} I think \textquotedblleft
such that $w=vu$\textquotedblright\ should be \textquotedblleft such that
$w=uv$\textquotedblright.

\item \textbf{Page 9:} \textquotedblleft Let $M$ be a graded right cancellable
monoid\textquotedblright\ should be \textquotedblleft Let $M$ be a graded
right-cancellable partial monoid generated by $M_{1}=\left\{  a_{1}%
,a_{2},a_{3},\ldots\right\}  $\textquotedblright. (The latter requirement is
needed to ensure that the condition $\max\operatorname*{supp}c\left(
w\right)  =\max\operatorname*{Last}\left(  w\right)  $ makes any sense; we
want $\max\operatorname*{Last}\left(  w\right)  $ to be a positive integer
after all.)

\item \textbf{Page 9:} In the displayed equation%
\begin{align*}
&  \#\left\{  w\in M\ \mid\ \ell\left(  w\right)  =n\text{ and }%
\max\operatorname*{Last}\left(  w\right)  =k\right\} \\
&  \leq\#\left\{  c\in\mathsf{Codes}\ \mid\ \left\vert c\left(  w\right)
\right\vert =n\text{ and }\max\operatorname*{supp}c\left(  w\right)
=k\right\}  ,
\end{align*}
replace both \textquotedblleft$c\left(  w\right)  $\textquotedblright s by
\textquotedblleft$c$\textquotedblright s. (There should not be any $w$ on the
right hand side.)

\item \textbf{Page 10, proof of Theorem 2.20:} In the equality sign%
\[
\sum_{\ell\left(  w\right)  =k}\sum_{i\in\operatorname*{Last}\left(  w\right)
}Y_{i}\left(  X_{w}\left(  f\right)  S_{w/i}\right)  =\sum_{\ell\left(
w\right)  =k}X_{w}\left(  f\right)  S_{w},
\]
you are tacitly using the fact that $X_{w}\left(  f\right)  $ is a constant
(since $\deg f=k=\ell\left(  w\right)  $) and thus can be moved out of the
$Y_{i}$. This is perhaps worth saying...

\item \textbf{Page 10, proof of Theorem 2.20:} In \textquotedblleft Writing
\newline$\mathsf{Codes}_{k,d}=\left\{  c\in\mathsf{Codes}\ \mid\ \max
\operatorname*{supp}c\leq d\text{ and }\left\vert c\left(  w\right)
\right\vert =k\right\}  $\textquotedblright, replace \textquotedblleft%
$c\left(  w\right)  $\textquotedblright\ by \textquotedblleft$c$%
\textquotedblright.

\item \textbf{Page 10, proof of Theorem 2.20:} \textquotedblleft for any
$\lambda\in\mathbb{Z}$\textquotedblright\ should be \textquotedblleft for any
nonzero $\lambda\in\mathbb{Z}$\textquotedblright.

\item \textbf{Page 10, proof of Theorem 2.20:} The argument you use to show
that $\left\{  S_{w}\ \mid\ w\in M_{k,d}\right\}  $ is a basis for
$\operatorname*{Pol}\nolimits_{d}^{\left(  k\right)  }$ is unnecessarily
complicated. There is a general fact (well-known for fields, easy for
$\mathbb{Z}$, not hard in general\footnote{For a proof, see, e.g., Proposition
II.5.2 in: \href{https://arxiv.org/abs/1605.04832v4}{Henri Lombardi \& Claude
Quitt\'{e}, \textit{Commutative algebra: Constructive methods},
arXiv:1605.04832v4}.}) that if $V$ is a free module of rank $m\in\mathbb{N}$
over a nontrivial commutative ring $\mathbf{k}$, then any family of $\leq m$
vectors in $V$ that spans $V$ must be a basis of $V$ and must furthermore
contain exactly $m$ vectors. Applying this to $\mathbf{k}=\mathbb{Z}$ and
$V=\operatorname*{Pol}\nolimits_{d}^{\left(  k\right)  }$ and
$m=\operatorname*{rank}\left(  \operatorname*{Pol}\nolimits_{d}\right)
=\left\vert \mathsf{Codes}_{k,d}\right\vert $ and to the family $\left(
S_{w}\right)  _{w\in M_{k,d}}\in V^{M_{k,d}}$ (which spans
$\operatorname*{Pol}\nolimits_{d}$ by (2.6), and consists of $\left\vert
M_{k,d}\right\vert \leq\left\vert \mathsf{Codes}_{k,d}\right\vert
=\operatorname*{rank}\left(  \operatorname*{Pol}\nolimits_{d}\right)  $
vectors), we conclude that this family $\left(  S_{w}\right)  _{w\in M_{k,d}}%
$must be a basis of $\operatorname*{Pol}\nolimits_{d}^{\left(  k\right)  }$
and must contain exactly $\left\vert \mathsf{Codes}_{k,d}\right\vert $ many
vectors, i.e., that $\left\vert M_{k,d}\right\vert =\left\vert \mathsf{Codes}%
_{k,d}\right\vert $. This is what you wanted to show.

A major advantage of this argument is that it works over any nontrivial
commutative ring $\mathbf{k}$, not just over $\mathbb{Z}$ or $\mathbb{Q}$.

\item \textbf{Page 11, \S 3:} \textquotedblleft The Lehmer code is the
bijective map $S_{\infty}\rightarrow\mathsf{Codes}$\textquotedblright%
\ $\rightarrow$ \textquotedblleft The Lehmer code is the bijective map
$\operatorname*{lcode}:S_{\infty}\rightarrow\mathsf{Codes}$\textquotedblright%
\ (since you then speak of $\operatorname*{lcode}\left(  w\right)  $).

\item \textbf{Page 12, \S 3.2:} \textquotedblleft staircase\textquotedblright%
\ $\rightarrow$ \textquotedblleft staircase Young diagram\textquotedblright.
(Otherwise, it is not clear how the $n$-tuple $\left(  n,n-1,\ldots,1\right)
$ suddenly has \textquotedblleft columns\textquotedblright.)

\item \textbf{Page 13, proof of Theorem 3.5:} In the first sentence, you are
using the fact that $\operatorname*{id}-\mathsf{R}_{1}$ is invertible on
$\operatorname*{Pol}\nolimits^{+}$ (so that the recursion (3.1) can be solved
for $\mathfrak{S}_{w}$ uniquely once $\mathfrak{S}_{1}=1$ is known).

\item \textbf{Page 13, proof of Theorem 3.5:} You talk of \textquotedblleft
pipes $i$ and $i+1$\textquotedblright. It should be explained that
\textquotedblleft pipe $k$\textquotedblright\ means the pipe that starts
(enters) in row $k$ at the left boundary of the diagram.

\item \textbf{Page 13, proof of Theorem 3.5:} In the last display, the
\textquotedblleft D\textquotedblright\ in \textquotedblleft D has no crosses
in row $i$\textquotedblright\ should be typeset in math mode ($D$, not D).

The same applies to \textquotedblleft nodes(D)\textquotedblright\ in the proof
of Theorem 4.3 (page 17).

\item \textbf{Page 14, definition of }$\mathsf{T}_{i}$\textbf{:} It should be
explained that \textquotedblleft$\dfrac{\mathsf{R}_{i+1}-\mathsf{R}_{i}}%
{x_{i}}$\textquotedblright\ means $x_{i}^{-1}\left(  \mathsf{R}_{i+1}%
-\mathsf{R}_{i}\right)  $, not $\left(  \mathsf{R}_{i+1}-\mathsf{R}%
_{i}\right)  x_{i}^{-1}$ (since operators don't commute).

\item \textbf{Page 14, definition of indexed forests:} The definition of an
indexed forest is somewhat ambiguous about whether the internal nodes are
labelled or unlabelled. As I understand, they should be unlabelled (i.e.,
strictly speaking, each $T_{i}$ is an unlabelled rooted planar binary tree;
its leaves of course obtain a natural labelling by the left-to-right reading order).

\item \textbf{Page 14, definition of the map }$\mathsf{c}$\textbf{:} Replace
\textquotedblleft$c_{i}=\left\{  v\mid\rho_{F}\left(  v\right)  =i\right\}
$\textquotedblright\ by \textquotedblleft$c_{i}=\#\left\{  v\mid\rho
_{F}\left(  v\right)  =i\right\}  $\textquotedblright.

\item \textbf{Page 14, definition of the map }$\mathsf{c}$\textbf{:} If I am
not mistaken, the bijectivity of the map $\mathsf{c}$ can be proved in a
simple way: As we know, the monoid $\mathsf{For}$ is generated by the
generators $\underline{1},\underline{2},\underline{3},\ldots$ which are
subject to the Thompson relations $\underline{i}\cdot\underline{j}%
=\underline{j}\cdot\underline{i+1}$ for all $i>j$. The latter relations allow
us to write each element $F\in\mathsf{For}$ in the form $\underline{i_{1}%
}\cdot\underline{i_{2}}\cdot\cdots\cdot\underline{i_{k}}$ for some positive
integers $i_{1}\leq i_{2}\leq\cdots\leq i_{k}$ (by successively rewriting
$\underline{i}\cdot\underline{j}$ as $\underline{j}\cdot\underline{i+1}$ when
$i>j$; it is not hard to check that this algorithm will always terminate,
because the sum of all numbers keep increasing but none of the numbers can
outgrow the label of the last leaf of a nontrivial tree in our forest $F$).
But it is easy to see that any positive integers $i_{1}\leq i_{2}\leq
\cdots\leq i_{k}$ satisfy $\mathsf{c}\left(  \underline{i_{1}}\cdot
\underline{i_{2}}\cdot\cdots\cdot\underline{i_{k}}\right)  =e_{i_{1}}e_{i_{2}%
}\cdots e_{i_{k}}$ in the monoid $\mathsf{Codes}$ (that is, the $p$-th entry
of the sequence $\mathsf{c}\left(  \underline{i_{1}}\cdot\underline{i_{2}%
}\cdot\cdots\cdot\underline{i_{k}}\right)  $ is the number of times $p$
appears in the $k$-tuple $\left(  i_{1},i_{2},\ldots,i_{k}\right)  $). Thus,
the map $\mathsf{c}:\mathsf{For}\rightarrow\mathsf{Codes}$ is bijective. This
also shows that each element $F\in\mathsf{For}$ can be written in the form
$\underline{i_{1}}\cdot\underline{i_{2}}\cdot\cdots\cdot\underline{i_{k}}$ for
a \textbf{unique} weakly increasing tuple of positive integers $i_{1}\leq
i_{2}\leq\cdots\leq i_{k}$. This, in turn, easily yields that the Thompson
relations give a presentation of $\mathsf{For}$ (that is, $\mathsf{For}$ is
the quotient of the free monoid by the Thompson relations).

\item \textbf{Page 14:} Replace \textquotedblleft The set
$\operatorname*{Last}\left(  w\right)  $ is identified with\textquotedblright%
\ by \textquotedblleft The set $\operatorname*{Last}\left(  F\right)  $ is
identified with\textquotedblright\ (you don't call anything $w$ here).

\item \textbf{Page 15, definition of the trimmed forest:} After
\textquotedblleft deleting the terminal node $v$ satisfying $\rho_{F}\left(
v\right)  =i$\textquotedblright, add \textquotedblleft and compressing the
leaves $i$ and $i+1$ into a single leaf (this causes all later leaves $j>i+1$
to be relabelled as $j-1$)\textquotedblright.

\item \textbf{Page 16, \S 4.2:} \textquotedblleft the relation $\mathsf{R}%
_{1}+\sum_{i\geq1}x_{i}\mathsf{T}_{i}=\operatorname*{id}$ from Corollary
3.2\textquotedblright\ should be \textquotedblleft the relation $\mathsf{R}%
_{1}+\sum_{i\geq1}x_{i}\mathsf{T}_{i}=\operatorname*{id}$ from Lemma
3.1\textquotedblright.

\item \textbf{Page 17, \S 4.2:} It would be nice to explain how exactly the
graphs correspond to forests. I don't see how to generalize Figure 6.

\item \textbf{Page 17, \S 4.3:} In the definition of $\mathsf{ThMon}%
^{\underline{m}}$, replace \textquotedblleft for $i>m$\textquotedblright\ by
\textquotedblleft for $i>j$\textquotedblright.

\item \textbf{Page 18, \S 5.2, definition of }$\mathsf{R}_{i}^{\infty}%
$\textbf{:} Replace \textquotedblleft$\mathsf{R}_{i}^{\infty}\left(  f\right)
=f\left(  x_{1},\ldots,x_{i-1},x_{i},0,0,\ldots\right)  $\textquotedblright%
\ by \textquotedblleft$\mathsf{R}_{i}^{\infty}\left(  f\right)  =f\left(
x_{1},\ldots,x_{i-2},x_{i-1},0,0,\ldots\right)  $\textquotedblright.

\item \textbf{Page 19, Definition 5.4:} \textquotedblleft Explicitly,
$\mathsf{B}_{i}$ vanishes outside of $\operatorname*{Pol}\nolimits_{i}%
$\textquotedblright\ should be \textquotedblleft Explicitly, $\mathsf{B}_{i}$
vanishes on all monomials outside of $\operatorname*{Pol}\nolimits_{i}%
$\textquotedblright.

\item \textbf{Page 19, Definition 5.4:} \textquotedblleft where $p_{j}>0$ or
$j=0$\textquotedblright\ should be \textquotedblleft where $j<i$ and $p_{j}>0$
or $j=0$\textquotedblright.

\item \textbf{Page 19, proof of Proposition 5.5:} The computation of $\left(
\mathsf{R}_{r+1}^{\infty}-\mathsf{R}_{r}^{\infty}\right)  f$ goes a bit too
fast for me. Here are some explanations I would make:

\begin{itemize}
\item The equality%
\begin{align*}
&  f\left(  x_{1},\ldots,x_{r-1},0^{j},x_{r},0,\ldots\right)  -f\left(
x_{1},\ldots,x_{r-1},0^{j+1},x_{r},0,\ldots\right) \\
&  =\mathsf{R}_{r}^{j}\mathsf{R}_{r+j+1}^{\infty}\left(  x_{r+j}%
-x_{r+j+1}\right)  \partial_{r+j}f
\end{align*}
follows from observing that
\begin{align*}
&  \left(  x_{r+j}-x_{r+j+1}\right)  \partial_{r+j}f\\
&  =f-\left(  f\text{ with the variables }x_{r+j}\text{ and }x_{r+j+1}\text{
swapped}\right)
\end{align*}
and%
\[
\mathsf{R}_{r}^{j}\mathsf{R}_{r+j+1}^{\infty}g=g\left(  x_{1},x_{2}%
,\ldots,x_{r-1},0^{j},x_{r},0^{\infty}\right)  \qquad\text{for each }%
g\in\operatorname*{Pol}%
\]
(where $0^{\infty}$ means an infinite sequence $0,0,0,\ldots$).

\item The equality
\[
\mathsf{R}_{r}^{j}\mathsf{R}_{r+j+1}^{\infty}\left(  x_{r+j}-x_{r+j+1}\right)
\partial_{r+j}f=x_{r}\mathsf{R}_{r}^{j}\mathsf{R}_{r+j+1}^{\infty}%
\partial_{r+j}f
\]
follows from the (easily proved) identities $\mathsf{R}_{r+j+1}^{\infty
}x_{r+j+1}=0$ and $\mathsf{R}_{r+j+1}^{\infty}x_{r+j}=x_{r+j}\mathsf{R}%
_{r+j+1}^{\infty}$ and $\mathsf{R}_{r}^{j}x_{r+j}=x_{r}\mathsf{R}_{r}^{j}$.

\item The equality%
\[
x_{r}\mathsf{R}_{r}^{j}\mathsf{R}_{r+j+1}^{\infty}\partial_{r+j}f=\left(
x_{r}\mathsf{R}_{r}^{j}\mathsf{R}_{r+j+1}^{\infty}\right)  \mathsf{D}_{r+j}f
\]
follows from the fact that the operator $\mathsf{R}_{r+j+1}^{\infty}$ (like
any $\mathsf{R}_{i}^{\infty}$) is idempotent.
\end{itemize}

\item \textbf{Page 20, proof of Proposition 5.6:} In the definition of
$A_{\left(  i_{1},\ldots,i_{k}\right)  }$, it should be pointed out that
$i_{0}$ is understood as $0$ (this becomes necessary when all entries of
$\left(  i_{1},\ldots,i_{k}\right)  $ are equal to $p$).

\item \textbf{Page 20, proof of Proposition 5.6:} Replace \textquotedblleft we
have $a_{k-\ell}\geq a_{k-\ell+1}\geq\cdots\geq a_{k}\geq p$\textquotedblright%
\ by \textquotedblleft we have $a_{\ell+1}\geq a_{\ell+2}\geq\cdots\geq
a_{k}\geq p$\textquotedblright.

\item \textbf{Page 21, proof of Proposition 5.7:} \textquotedblleft We can
then use Theorem 2.20\textquotedblright\ $\rightarrow$ \textquotedblleft We
can then use Proposition 2.16\textquotedblright.

\item \textbf{Page 21, proof of Lemma 5.10:} Why does it suffice to show that
\textquotedblleft$\mathsf{D}_{i}\mathsf{R}_{j}\mathfrak{F}_{a}=0$ for all $i$,
except at most one for which $\mathsf{D}_{i}\mathsf{R}_{j}\mathfrak{F}%
_{a}=\mathfrak{F}_{b}$ for some $b\in\mathsf{Winc}$\textquotedblright?
Shouldn't it also be shown that $i$ is $\geq$ to the last entry of $b$, so
that Theorem 5.3 can be applied?

(Pedantic remark: The \textquotedblleft$\mathsf{Winc}$\textquotedblright\ here
should be a \textquotedblleft$\mathsf{WInc}$\textquotedblright.)

\item \textbf{Page 22, Slide Kostka matrix:} In the definition of
$E_{i}\left(  a\right)  $, what does \textquotedblleft$\in\left\{
0,1\right\}  =0$\textquotedblright\ mean?
\end{enumerate}


\end{document}