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\ihead{Errata to ``On the straightening law...''}
\ohead{\today}
\begin{document}

\begin{center}
\textbf{On the straightening law for minors of a matrix}

\textit{Richard G. Swan}

\texttt{\href{http://www.math.uchicago.edu/~swan/strLaw.pdf}{http://www.math.uchicago.edu/\symbol{126}%
swan/strLaw.pdf}}

version of 18 February 2003

\textbf{Errata and addenda by Darij Grinberg}

\bigskip
\end{center}

I will refer to the results appearing in the preprint \textquotedblleft On the
straightening law for minors of a matrix\textquotedblright\ by the numbers
under which they appear in this preprint (specifically, in its version of 18
February 2003, published on \newline%
\texttt{\href{http://www.math.uchicago.edu/~swan/strLaw.pdf}{http://www.math.uchicago.edu/\symbol{126}%
swan/strLaw.pdf}}).

\setcounter{section}{10}

\section{Errata}

\begin{itemize}
\item \textbf{Page 1:} The name \textquotedblleft Doubilet\textquotedblright%
\ is misspelt as \textquotedblleft Doubillet\textquotedblright\ twice (once in
the abstract, and once again in \S 1).,

\item \textbf{Page 2, proof of Lemma 2.3:} Replace \textquotedblleft etc.
,each\textquotedblright\ by \textquotedblleft etc., each\textquotedblright.

\item \textbf{Page 2, proof of Theorem 2.6:} The \textquotedblleft$\sigma
$\textquotedblright\ comes out of the blue in the proof, as there is no
$\sigma$ in the statement of the theorem. To clarify where it comes from, I
suggest adding the following sentence at the beginning of the proof:
\textquotedblleft By Corollary 2.5, we can WLOG assume that $X$ is a
permutation matrix $P\left(  \sigma^{-1}\right)  $ for some $\sigma
\in\mathcal{S}_{n}$.\textquotedblright.

\item \textbf{Page 3, \S 3:} Replace \textquotedblleft say that S is
good\textquotedblright\ by \textquotedblleft say that $S$ is
good\textquotedblright. (I have replaced the text-\textquotedblleft
S\textquotedblright\ by a math-\textquotedblleft$S$\textquotedblright\ here.)

\item \textbf{Page 3, Theorem 3.1:} It would be helpful to clarify that some
of the $\left(  A_{i},B_{i}\right)  $ may be equal.

\item \textbf{Page 3, proof of Theorem 3.1:} Replace \textquotedblleft%
$B_{i}\leq B$,and\textquotedblright\ by \textquotedblleft$B_{i}\leq B$,
and\textquotedblright.

\item \textbf{Page 3, proof of Theorem 3.1:} Replace \textquotedblleft all
other terms\textquotedblright\ by \textquotedblleft all other nonzero
terms\textquotedblright.

\item \textbf{Page 4, proof of Theorem 4.1:} This proof is confusing due to
some of its parts being out of order. The determinant $Y\left(  P\mid
Q\right)  $ is not well-defined before the sets $I$ and $J$ are totally
ordered; I even would not call $Y$ a \textquotedblleft square
matrix\textquotedblright\ until the indexing set $I$ for its rows and the
indexing set $J$ for its columns have been identified. I suggest modifying the
proof as follows:

\begin{itemize}
\item At the beginning of the proof, add the following sentence:
\textquotedblleft We WLOG assume that $\left\vert S^{\prime}\right\vert
=\left\vert T^{\prime}\right\vert $ and $\left\vert S^{\prime\prime
}\right\vert =\left\vert T^{\prime\prime}\right\vert $, as otherwise the left
hand side is $0$.\textquotedblright.

\item Move the second paragraph of the proof (the paragraph that begins with
\textquotedblleft Order $I$ by setting\textquotedblright\ and ends with
\textquotedblleft and similarly for subsets of $J$\textquotedblright) to
before the sentence that begins with \textquotedblleft Define a square matrix
$Y$\textquotedblright.

\item Before the sentence that begins with \textquotedblleft Define a square
matrix $Y$\textquotedblright, add the following sentence: \textquotedblleft
Now the sets $I$ and $J$ are totally ordered. Let $k=\left\vert I\right\vert
=\left\vert J\right\vert $. Then, we have order-preserving bijections
$I\rightarrow\left\{  1,2,\ldots,k\right\}  $ and $J\rightarrow\left\{
1,2,\ldots,k\right\}  $. Use these bijections to identify $I$ and $J$ with
$\left\{  1,2,\ldots,k\right\}  $. For every subset $Q$ of $\left\{
1,2,\ldots,k\right\}  $, we will write $\widetilde{Q}$ for the complement
$\left\{  1,2,\ldots,k\right\}  -Q$.\textquotedblright.
\end{itemize}

Once this is done, $Y$ actually becomes a honest square matrix (of size
$k\times k$).

\item \textbf{Page 4, proof of Theorem 4.1:} Replace \textquotedblleft$\left(
S^{\prime}\mid T^{\prime}\right)  \left(  S^{\prime\prime}\mid T^{\prime
\prime}\right)  =Y\left\{  I^{\prime}\mid J^{\prime}\right\}  $%
\textquotedblright\ by \textquotedblleft$\left(  S^{\prime}\mid T^{\prime
}\right)  \left(  S^{\prime\prime}\mid T^{\prime\prime}\right)  =\pm Y\left\{
I^{\prime}\mid J^{\prime}\right\}  $\textquotedblright. (At least I don't see
a reason why the $\pm$ must always be a $+$. Maybe it is?)

\item \textbf{Page 5, proof of Lemma 4.2:} Add \textquotedblleft WLOG assume
that $\left\vert K\right\vert =\left\vert I^{\prime}\right\vert $ (since
otherwise, $\varphi\left(  K\right)  \neq S^{\prime}$ is
obvious).\textquotedblright\ after the first sentence of the proof.

\item \textbf{Page 5, proof of Lemma 4.2:} Remove \textquotedblleft,
$S^{\prime}=\left\{  s_{1}<\cdots<s_{p}\right\}  $\textquotedblright\ (as this
is unnecessary).

\item \textbf{Page 5, proof of Lemma 4.2:} The argument \textquotedblleft But
$i_{\nu}$ lies in $I^{\prime}$ so $k_{\nu}$ is in $I^{\prime\prime}$ and
therefore $k_{\nu}>i_{\nu}$\textquotedblright\ goes a bit too fast for me. I
suggest adding a few details, e.g., as follows:

\textquotedblleft Recall that $\varphi$ is injective on $I^{\prime}$ and on
$K$. Thus, from $I^{\prime}=\left\{  i_{1}<\cdots<i_{p}\right\}  $ and
$K=\left\{  k_{1}<\cdots<k_{p}\right\}  $, we obtain $\varphi\left(
I^{\prime}\right)  =\left\{  \varphi\left(  i_{1}\right)  <\cdots
<\varphi\left(  i_{p}\right)  \right\}  $ and $\varphi\left(  K\right)
=\left\{  \varphi\left(  k_{1}\right)  <\cdots<\varphi\left(  k_{p}\right)
\right\}  $. Hence, $\varphi\left(  K\right)  =S^{\prime}=\varphi\left(
I^{\prime}\right)  $ shows that $\varphi\left(  k_{\nu}\right)  =\varphi
\left(  i_{\nu}\right)  $. Since $k_{\nu}\neq i_{\nu}$ and $i_{\nu}\in
I^{\prime}$, this entails $k_{\nu}\in I^{\prime\prime}$ and therefore $k_{\nu
}>i_{\nu}$\textquotedblright.

\item \textbf{Page 5, \S 5:} A nitpick: Replace \textquotedblleft whose
entries are indeterminates\textquotedblright\ by \textquotedblleft whose
entries are distinct indeterminates\textquotedblright.

\item \textbf{Page 5, Theorem 5.3:} In this theorem, you probably want to
state that two standard monomials which only differ by factors of the form
$\left(  \varnothing,\varnothing\right)  $ are considered to be identical.
(You can always add $\left(  \varnothing,\varnothing\right)  $ to the end of a
standard monomial without changing the value of this monomial.)

\item \textbf{Page 6, proof of Theorem 5.3:} Replace all appearances of
\textquotedblleft leading form\textquotedblright\ by \textquotedblleft leading
term\textquotedblright\ (or is \textquotedblleft leading
form\textquotedblright\ really a synonym for \textquotedblleft leading
term\textquotedblright?).

\item \textbf{Page 6, proof of Theorem 5.3:} Add a period before
\textquotedblleft Similarly, if\textquotedblright.

\item \textbf{Page 6, proof of Theorem 5.3:} After \textquotedblleft$z\left(
B\right)  =z_{b_{1}}^{\left(  1\right)  }\cdots z_{b_{p}}^{\left(  p\right)
}$\textquotedblright, add \textquotedblleft(if $N\geq\left\vert A\right\vert
=\left\vert B\right\vert $)\textquotedblright.

\item \textbf{Page 6, References:} \textquotedblleft
Desarmenian\textquotedblright\ $\rightarrow$ \textquotedblleft
Desarmenien\textquotedblright\ in [4].
\end{itemize}


\end{document}