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\begin{center}
\textbf{The Littlewood-Richardson rule, and related combinatorics}

\textit{Marc A. A. van Leeuwen}

version of 19 May 2013

\textbf{Errata by Darij Grinberg - I}

\bigskip
\end{center}

The following text is an annotation to Marc A. A. van Leeuwen's paper
\textquotedblleft The Littlewood-Richardson rule, and related
combinatorics\textquotedblright\ in its version of 20 March 2012.

This annotation contains short comments as well as corrections of the
occasional mistake (or what I believe to be mistakes).

\rule{17cm}{0.3mm}

\textbf{Page 2, \S 1.1:} ``the the action''\ $\rightarrow$ ``the action''.

\rule{17cm}{0.3mm}

\textbf{Page 2, \S 1.2:} Here, you write: ``The Schur polynomials of degree
$d$ form yet another $\mathbb{Z}$-basis of $\Lambda_{n}$ parametrized by
$\mathcal{P}_{d,n}$.''\ The $\Lambda_{n}$ here should be replaced by
$\Lambda_{n}^{d}$.

\rule{17cm}{0.3mm}

\textbf{Page 4, \S 1.2:} In ``the fact that $\left\{  m_{\lambda}\left(
n\right)  \ \mid\ \lambda\in\mathcal{P}_{d,n}\right\}  $ is a $\mathbb{Z}%
$-basis of $\Lambda_{n}$'', the $\Lambda_{n}$ again should be replaced by
$\Lambda_{n}^{d}$.

\rule{17cm}{0.3mm}

\textbf{Page 4, \S 1.3:} You claim that \textquotedblleft this representation
is not unique in general (although in some cases it is, for instance when the
sets of rows and columns meeting $D$ are both initial intervals of
$\mathbb{N}$)\textquotedblright. This is not literally correct, since the
representation is not unique when $D$ is the empty skew diagram, whereas it is
clear that the sets of rows and columns meeting the empty skew diagram are
both initial intervals of $\mathbb{N}$. However, this is the only exception.
In fact, something more general holds: If the set of rows meeting a skew
diagram $D$ is an initial interval of $\mathbb{N}$, and the set of columns
meeting $D$ contains $0$, then the representation of $D$ as a difference of
two Young diagrams is unique. (Of course, the same holds with
\textquotedblleft rows\textquotedblright\ and \textquotedblleft
columns\textquotedblright\ switched. These two cases, however, don't cover all
cases where the difference representation is unique; and in general, whether
or not the representation is unique is not determined by the sets of rows and
of columns meeting $D$.)

\rule{17cm}{0.3mm}

\textbf{Page 6, Proposition 1.4.3:} This proposition is left unproved here,
and indeed the proof is not exactly deep -- but I would not call it trivial
either. I have spelled out the proof in detail in Exercise 2.9.20 (a) in:

\begin{quote}
\href{https://arxiv.org/abs/1409.8356v7}{Darij Grinberg, Victor Reiner,
\textit{Hopf algebras in combinatorics}, arXiv:1409.8356v7}
\end{quote}

\noindent(specifically, the claim of Proposition 1.4.3 is the equivalence of
Assertions $\mathcal{F}^{\left(  \kappa\right)  }$ and $\mathcal{G}^{\left(
\kappa\right)  }$ in that exercise, since Assertion $\mathcal{G}^{\left(
\kappa\right)  }$ is saying that $T$ is $\kappa$-dominant, whereas Assertion
$\mathcal{F}^{\left(  \kappa\right)  }$ is saying that $\alpha\in\mathcal{P}$
throughout the test formulated in the proposition).

\rule{17cm}{0.3mm}

\textbf{Page 8, \S 1.5:} ``another a valid reading order''\ should be
``another valid reading order''.

\rule{17cm}{0.3mm}

\textbf{Page 9, \S 2.1:} You haven't defined the relation \textquotedblleft%
$\subseteq$\textquotedblright\ in \textquotedblleft$\left(  \mathcal{P}%
,\subseteq\right)  $\textquotedblright. (Of course, the definition is easy: If
$\lambda$ and $\mu$ are two partitions, then we set $\lambda\subseteq\mu$ if
and only if every $i\geq0$ satisfies $\lambda_{i}\leq\mu_{i}$, where $\lambda$
and $\mu$ have been written as $\lambda=\left(  \lambda_{0},\lambda
_{1},\lambda_{2},...\right)  $ and $\mu=\left(  \mu_{0},\mu_{1},\mu
_{2},...\right)  $, respectively. In other words, if $\lambda$ and $\mu$ are
two partitions, then $\lambda\subseteq\mu$ if and only if $Y\left(
\lambda\right)  \subseteq Y\left(  \mu\right)  $.)

\rule{17cm}{0.3mm}

\textbf{Page 9, \S 2.2:} On the first line of \S 2.2, remove the word
``of''\ from the formulation ``Among the skew shapes $\lambda/\mu$ of with
$\left\vert \lambda/\mu\right\vert =2$''.

\rule{17cm}{0.3mm}

\textbf{Page 10, \S 2.2:} Shortly before Definition 2.2.1, you write:

``It follows that if values $\lambda^{\left[  i,j\right]  }$ are prescribed
for indices $\left[  i,j\right]  $ traversing some ``lattice path''\ going
from $\left[  k,m\right]  $ to $\left[  l,n\right]  $ (a zig-zag path in which
at each step either $i$ or $j$ increases by $1$) by any skew standard tableau,
then there is a unique way to extend these values to tableau switching family
on $I\times J$''.

First, ``tableau switching family''\ should be ``a tableau switching family''\ here.

On a less trivial note, it took me a while to figure out why this extension
exists and is unique. A posteriori, the difficulties were illusory: Since
there are many orders in which we can compute the values of this extension, my
intuition suggested that there can also be many different results and perhaps
even some \textquotedblleft holonomy\textquotedblright\ that would need to be
proven trivial. In truth, the proof is easy and requires no such work:

\begin{itemize}
\item First one proves the \textbf{existence} of the tableau switching family
$\left(  \lambda^{\left[  i,j\right]  }\right)  _{k\leq i\leq l;\ m\leq j\leq
n}$ with given values on a given lattice path from $\left[  k,m\right]  $ to
$\left[  l,n\right]  $. This can be done by recursively computing the entries
$\lambda^{\left[  i,j\right]  }$ of the family as follows: Start with the
given entries on the lattice path (which are already known). Then, compute all
the entries $\lambda^{\left[  i,j\right]  }$ above the lattice path (in the
order in which they become computable: until all these entries have been
computed, there is always at least one uncomputed $\lambda^{\left[
i,j\right]  }$ whose neighbors $\lambda^{\left[  i+1,j\right]  }%
,\ \lambda^{\left[  i,j+1\right]  },\ \lambda^{\left[  i+1,j+1\right]  }$ are
all computed already, and so we pick such an entry and compute it). Finally,
compute all the entries $\lambda^{\left[  i,j\right]  }$ below the lattice
path (in an analogous way). Thus, at least one tableau switching family
$\left(  \lambda^{\left[  i,j\right]  }\right)  _{k\leq i\leq l;\ m\leq j\leq
n}$ exists that has the given values on our given lattice path.

\item To prove its \textbf{uniqueness}, we must show that if $\left(
\lambda^{\left[  i,j\right]  }\right)  _{k\leq i\leq l;\ m\leq j\leq n}$ and
$\left(  \mu^{\left[  i,j\right]  }\right)  _{k\leq i\leq l;\ m\leq j\leq n}$
are two tableau switching families that agree on a given lattice path from
$\left[  k,m\right]  $ to $\left[  l,n\right]  $, then they agree everywhere.
This, again, is proved recursively, just as in the proof of existence: We are
given that $\lambda^{\left[  i,j\right]  }=\mu^{\left[  i,j\right]  }$ for all
$\left(  i,j\right)  $ on the lattice path. Next, we recursively prove that
$\lambda^{\left[  i,j\right]  }=\mu^{\left[  i,j\right]  }$ for all points
$\left(  i,j\right)  $ above the lattice path (again recursively, e.g., by
strong induction on $-i-j$, so that the induction hypothesis already shows
that $\lambda^{\left[  i+1,j\right]  }=\mu^{\left[  i+1,j\right]  }$ and
$\lambda^{\left[  i,j+1\right]  }=\mu^{\left[  i,j+1\right]  }$ and
$\lambda^{\left[  i+1,j+1\right]  }=\mu^{\left[  i+1,j+1\right]  }$, and thus
the uniqueness of the tableau-switching rule allows us to conclude that
$\lambda^{\left[  i,j\right]  }=\mu^{\left[  i,j\right]  }$). Finally, we
prove that $\lambda^{\left[  i,j\right]  }=\mu^{\left[  i,j\right]  }$ for all
points $\left(  i,j\right)  $ below the lattice path (similarly). Thus,
$\lambda^{\left[  i,j\right]  }=\mu^{\left[  i,j\right]  }$ is proved for all
$\left(  i,j\right)  $; this completes the proof of uniqueness.
\end{itemize}

\rule{17cm}{0.3mm}

\textbf{Page 11, \S 2.2:} There is a typo in the paragraph succeeding Theorem
2.2.2: ``exactly that of in inward \textit{jeu de taquin} slide''\ should be
``exactly that of an inward \textit{jeu de taquin} slide''.

\rule{17cm}{0.3mm}

\textbf{Page 11, \S 2.2:} I think it would add value to the exposition in the
paper if the notion of a jeu de taquin slide would be defined (particularly
the \textquotedblleft inward slide into\textquotedblright\ and
\textquotedblleft outward slide into\textquotedblright\ terminology). While it
is possible to define jeu de taquin slides through tableau switching, I don't
think you make it very clear how to do that, and you do use \textquotedblleft
sliding\textquotedblright\ metaphors further in the text. You kind-of define
jeu de taquin slides in
\href{http://wwwmathlabo.univ-poitiers.fr/~maavl/pdf/pictures.pdf}{your
\texttt{pictures.pdf} paper}, but for some reason you call them
\textquotedblleft glissements\textquotedblright\ there...

\rule{17cm}{0.3mm}

\textbf{Page 12, \S 2.3:} In \textquotedblleft indicate the first second and
third\textquotedblright, add a comma after \textquotedblleft
first\textquotedblright.

\rule{17cm}{0.3mm}

\textbf{Page 12, \S 2.3:} In the Young tableau one line above Proposition
2.3.2, why are the entries $1,2,3$ called rather than $0,1,2$ ?

\rule{17cm}{0.3mm}

\textbf{Page 13, \S 2.3:} \textquotedblleft If $t$ lies a
diagonal\textquotedblright\ should be \textquotedblleft If $t$ lies on a
diagonal\textquotedblright.

\rule{17cm}{0.3mm}

\textbf{Page 13, \S 2.3:} One line above Proposition 2.3.3, \textquotedblleft
consequence proposition 2.2.5\textquotedblright\ should be \textquotedblleft
consequence of proposition 2.2.5\textquotedblright.

\rule{17cm}{0.3mm}

\textbf{Page 13, proof of Theorem 2.4.1:} \textquotedblleft between those of
$p$ and $q$\textquotedblright\ should be \textquotedblleft between those of
$p$ and $r$\textquotedblright.

\rule{17cm}{0.3mm}

\textbf{Page 14, Proposition 2.4.2:} This is worded quite ambiguously: Does
\textquotedblleft any\textquotedblright\ means \textquotedblleft
some\textquotedblright\ or \textquotedblleft all\textquotedblright? Is the
tableau $T$ supposed to be skew or straight? standard or just semistandard?

In truth, several of these options work and are being used later! Thus, I
suggest replacing Proposition 2.4.2 by the following:

\begin{statement}
\textbf{2.4.2. Proposition.} Let $S_{1}$ and $S_{2}$ be two skew semistandard
tableaux of the same shape $\lambda/\mu$. Then, the following three statements
are equivalent:

\begin{enumerate}
\item[\textbf{(1)}] The tableaux $S_{1}$ and $S_{2}$ are dual equivalent.

\item[\textbf{(2)}] There is a standard tableau $T$ of shape $\mu$ such that
if we set $X\left(  T,S_{i}\right)  =\left(  S_{i}^{\prime},T_{i}^{\prime
}\right)  $ for $i=1,2$, then $T_{1}^{\prime}=T_{2}^{\prime}$.

\item[\textbf{(3)}] For any skew semistandard tableau $T$ of any shape
$\mu/\nu$, if we set $X\left(  T,S_{i}\right)  =\left(  S_{i}^{\prime}%
,T_{i}^{\prime}\right)  $ for $i=1,2$, then $T_{1}^{\prime}=T_{2}^{\prime}$.
\end{enumerate}
\end{statement}

The first three sentences of the proof given for Proposition 2.4.2 are proving
the implication \textbf{(2)} $\Longrightarrow$ \textbf{(1)} (and the notion of
\textquotedblleft reverse dual equivalence\textquotedblright\ is used for
statement \textbf{(2)} there). The last sentence of the proof is proving the
implication \textbf{(1)} $\Longrightarrow$ \textbf{(3)}. The implication
\textbf{(1)} $\Longrightarrow$ \textbf{(3)} (and the notion of
\textquotedblleft reverse dual equivalence\textquotedblright\ is used for
statement \textbf{(3)} there). Combining these implications with the obvious
implication \textbf{(3)} $\Longrightarrow$ \textbf{(2)}, we conclude that all
three statements are equivalent, and so the updated Proposition 2.4.2 follows.

In the next paragraph, the implication \textbf{(2)} $\Longrightarrow$
\textbf{(1)} is used (since any application of successive inward slides into
$S_{1}$ and $S_{2}$ until obtaining a straight-shaped Young tableau amounts to
computing $X\left(  T,S_{i}\right)  $ for a standard tableau $T$ of shape
$\mu$).

\rule{17cm}{0.3mm}

\textbf{Page 17, Proposition 3.1.1:} Remove one of the words ``to''\ from
``associating to $w$ to the sequence''.

\rule{17cm}{0.3mm}

\textbf{Page 17:} When you say ``connected component'', it might be useful to
point out that these words mean a connected component of the
\textbf{undirected} graph obtained by forgetting the directions of the edges
in the coplactic graph.

\rule{17cm}{0.3mm}

\textbf{Page 21, proof of Corollary 3.3.3:} There is a ``the the''\ typo here.

\rule{17cm}{0.3mm}

\textbf{Page 28, \S 4.1:} Typo: ``estabishes''.

\rule{17cm}{0.3mm}
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