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\ihead{Math 701, Fall 2026, Lecture 10, version \today}
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\begin{document}
\section*{Math 701 Fall 2026, Lecture 10: The Murnaghan--Nakayama rule}

\textbf{website:}
\texttt{\href{https://www.cip.ifi.lmu.de/~grinberg/t/26fs}{\texttt{https://www.cip.ifi.lmu.de/\symbol{126}%
grinberg/t/26fs/}}}

(GPT-6 was used to create examples and to polish and detail some of the
arguments below. \textbf{The current version contains unverified GPT output.})
\medskip

\setcounter{section}{1}\setcounter{subsection}{9}\setcounter{subsubsection}{0}

\subsection{The Murnaghan--Nakayama rule}

So we can multiply Schur polynomials by $e_{n}$ and by $h_{n}$. What about
multiplying them by $p_{n}$ ?

Again, we recall $s_{\mu}=a_{\mu+\rho}/a_{\rho}$ for an $N$-partition
$\mu$, and we try to compute $p_{n}a_{\mu+\rho}$, or, in general,
$p_{n}a_{\alpha}$. Here and throughout this lecture, $n$ is a positive
integer. Recall that
\[
a_{\alpha}=\det\left(\left(x_{i}^{\alpha_{j}}\right)_{i,j\in[N]}\right)
\qquad\text{and}\qquad p_{n}=\sum_{k=1}^{N}x_{k}^{n}.
\]
We will use the following lemma about determinants:

\begin{lemma}
[Moving column lemma]\label{lem.det.moving-column}Let $\left(  b_{i,j}\right)  _{i,j\in\left[  N\right]  }$
be an $N\times N$-matrix over some commutative ring $R$, and let $c_{1}%
,c_{2},\ldots,c_{N}$ be $N$ elements of $R$. Then,%
\[
\left(  \sum_{k=1}^{N}c_{k}\right)  \det\left(  b_{i,j}\right)  _{i,j\in
\left[  N\right]  }=\sum_{k=1}^{N}\det\left(  b_{i,j}c_{i}^{\left[
j=k\right]  }\right)  _{i,j\in\left[  N\right]  }.
\]
Here, we are using the Iverson bracket notation: If $\mathcal{A}$ is a
statement, then $\left[  \mathcal{A}\right]  $ means the truth value of
$\mathcal{A}$:
\[
\left[\mathcal{A}\right]:=
\begin{cases}
1, & \text{if }\mathcal{A}\text{ is true};\\
0, & \text{if }\mathcal{A}\text{ is false}.
\end{cases}
\]
So the factor $c_{i}^{\left[  j=k\right]  }$ is $c_{i}^{0}=1$ if $j\neq k$
and is $c_{i}^{1}=c_{i}$ if $j=k$. Thus, the matrix $\left(  b_{i,j}%
c_{i}^{\left[  j=k\right]  }\right)  _{i,j\in\left[  N\right]  }$ is the
original matrix $\left(  b_{i,j}\right)  _{i,j\in\left[  N\right]  }$ with its
$k$-th column entrywise multiplied by the vector $\left(  c_{1},c_{2}%
,\ldots,c_{N}\right)  ^{T}$.
\end{lemma}

\Needspace{10\baselineskip}
\begin{example}
\label{exa.det.moving-column}Let $N=3$. Here the claim of
Lemma \ref{lem.det.moving-column} is
\begin{align*}
& (c_{1}+c_{2}+c_{3})
\det\begin{pmatrix}
a_{1,1}&a_{1,2}&a_{1,3}\\
a_{2,1}&a_{2,2}&a_{2,3}\\
a_{3,1}&a_{3,2}&a_{3,3}
\end{pmatrix}\\[4pt]
&\quad=\det\begin{pmatrix}
c_{1}a_{1,1}&a_{1,2}&a_{1,3}\\
c_{2}a_{2,1}&a_{2,2}&a_{2,3}\\
c_{3}a_{3,1}&a_{3,2}&a_{3,3}
\end{pmatrix}
+\det\begin{pmatrix}
a_{1,1}&c_{1}a_{1,2}&a_{1,3}\\
a_{2,1}&c_{2}a_{2,2}&a_{2,3}\\
a_{3,1}&c_{3}a_{3,2}&a_{3,3}
\end{pmatrix}\\[4pt]
&\qquad+\det\begin{pmatrix}
a_{1,1}&a_{1,2}&c_{1}a_{1,3}\\
a_{2,1}&a_{2,2}&c_{2}a_{2,3}\\
a_{3,1}&a_{3,2}&c_{3}a_{3,3}
\end{pmatrix}.
\end{align*}
Each determinant on the right changes just one column; the three terms
account for the three possible choices of that column.
\end{example}

\begin{proof}[Proof of Lemma \ref{lem.det.moving-column}.]
\allowdisplaybreaks[1]
For each $k\in\left[  N\right]  $, we have%
\begin{align*}
&  \det\left(  b_{i,j}c_{i}^{\left[  j=k\right]  }\right)  _{i,j\in\left[
N\right]  }\\
&  =\sum_{\sigma\in S_{N}}\left(  -1\right)  ^{\sigma}\prod_{i=1}%
^{N}b_{i,\sigma\left(  i\right)  }c_{i}^{\left[  \sigma\left(  i\right)
=k\right]  }\\
&  =\sum_{\sigma\in S_{N}}\left(  -1\right)  ^{\sigma}\prod_{i=1}%
^{N}b_{i,\sigma\left(  i\right)  }\underbrace{\prod_{i=1}^{N}c_{i}^{\left[
\sigma\left(  i\right)  =k\right]  }}_{=c_{\sigma^{-1}\left(  k\right)  }}\\
&  =\sum_{\sigma\in S_{N}}\left(  -1\right)  ^{\sigma}c_{\sigma^{-1}\left(
k\right)  }\prod_{i=1}^{N}b_{i,\sigma\left(  i\right)  }.
\end{align*}
Summing this over all $k\in[N]$, we find%
\begin{align*}
&  \sum_{k=1}^{N}\det\left(  b_{i,j}c_{i}^{\left[  j=k\right]  }\right)
_{i,j\in\left[  N\right]  }\\
&  =\sum_{k=1}^{N}\ \ \sum_{\sigma\in S_{N}}\left(  -1\right)  ^{\sigma
}c_{\sigma^{-1}\left(  k\right)  }\prod_{i=1}^{N}b_{i,\sigma\left(  i\right)
}\\
&  =\sum_{\sigma\in S_{N}}\left(  -1\right)  ^{\sigma}\underbrace{\left(
\sum_{k=1}^{N}c_{\sigma^{-1}\left(  k\right)  }\right)  }_{\substack{=c_{1}%
+c_{2}+\cdots+c_{N}\\\text{(since }\sigma\text{ is a permutation)}}%
}\prod_{i=1}^{N}b_{i,\sigma\left(  i\right)  }\\
&  =\sum_{\sigma\in S_{N}}\left(  -1\right)  ^{\sigma}\left(  c_{1}%
+c_{2}+\cdots+c_{N}\right)  \prod_{i=1}^{N}b_{i,\sigma\left(  i\right)  }\\
&  =\left(  c_{1}+c_{2}+\cdots+c_{N}\right)  \underbrace{\left(  \sum
_{\sigma\in S_{N}}\left(  -1\right)  ^{\sigma}\prod_{i=1}^{N}b_{i,\sigma
\left(  i\right)  }\right)  }_{=\det\left(  b_{i,j}\right)  _{i,j\in\left[
N\right]  }}\\
&  =\left(  c_{1}+c_{2}+\cdots+c_{N}\right)  \det\left(  b_{i,j}\right)
_{i,j\in\left[  N\right]  },
\end{align*}
as desired. The same identity holds when $N=0$: both sides are zero.
\end{proof}

So let us apply this to compute $p_{n}a_{\alpha}$, where $n$ is a positive
integer and $\alpha=\left(  \alpha_{1},\alpha_{2},\ldots,\alpha_{N}\right)
\in\mathbb{N}^{N}$ is an $N$-tuple. We have%
\begin{align*}
p_{n}a_{\alpha}  &  =\left(  x_{1}^{n}+x_{2}^{n}+\cdots+x_{N}^{n}\right)
\det\left(  x_{i}^{\alpha_{j}}\right)  _{i,j\in\left[  N\right]  }\\
&  =\sum_{k=1}^{N}\det\left(  x_{i}^{\alpha_{j}}\left(  x_{i}^{n}\right)
^{\left[  j=k\right]  }\right)  _{i,j\in\left[  N\right]  }%
\ \ \ \ \ \ \ \ \ \ \left(  \text{by Lemma \ref{lem.det.moving-column}}\right) \\
&  =\sum_{k=1}^{N}\det\left(  x_{i}^{\alpha_{j}+n\left[  j=k\right]  }\right)
_{i,j\in\left[  N\right]  }.
\end{align*}


Let $\overrightarrow{e}_{1},\overrightarrow{e}_{2},\ldots,\overrightarrow{e}%
_{N}\in\mathbb{N}^{N}$ be the standard basis vectors of the $\mathbb{Z}%
$-module $\mathbb{Z}^{N}$; that is, $\overrightarrow{e}_{i}$ has a $1$ in its
$i$-th position and zeroes everywhere else. Then, the exponent $\alpha
_{j}+n\left[  j=k\right]  $ can be rewritten as $\left(  \alpha
+n\overrightarrow{e}_{k}\right)  _{j}$ (since both are $\alpha_{j}$ when
$j\neq k$, and are $\alpha_{j}+n$ when $j=k$). So the above equality rewrites
as%
\begin{equation}
\label{eq.schur.power-alternant}
p_{n}a_{\alpha}=\sum_{k=1}^{N}\underbrace{\det\left(  x_{i}^{\left(
\alpha+n\overrightarrow{e}_{k}\right)  _{j}}\right)  _{i,j\in\left[  N\right]
}}_{=a_{\alpha+n\overrightarrow{e}_{k}}}=\sum_{k=1}^{N}a_{\alpha
+n\overrightarrow{e}_{k}}.
\end{equation}
Setting $\alpha=\mu+\rho$ (where $\mu$ is an $N$-partition) and dividing this
by $a_{\rho}$ should give us a formula for $p_{n}s_{\mu}$ as a sum of
Schur polynomials. As in Lecture 8, division by $a_{\rho}$ here means
cancelling this regular element from a polynomial identity; no assumption
that $K$ is a field is needed.

\Needspace{10\baselineskip}
\begin{example}
\label{exa.schur.power2-221}Let us compute $p_{2}s_{(2,2,1)}$ for $N=3$. We
have
\[
s_{\left(  2,2,1\right)  }=a_{\left(  2,2,1\right)  +\rho}/a_{\rho}=a_{\left(
4,3,1\right)  }/a_{\left(  2,1,0\right)  },
\]
so we must find $p_{2}a_{\left(  4,3,1\right)  }$. The above formula gives us%
\begin{align*}
p_{2}a_{\left(  4,3,1\right)  } &  =a_{\left(  6,3,1\right)  }%
+\underbrace{a_{\left(  4,5,1\right)  }}_{\substack{=-a_{\left(  5,4,1\right)
}\\\text{(since alternants}\\\text{are alternating)}}}+\underbrace{a_{\left(
4,3,3\right)  }}_{\substack{=0\\\text{(since alternants}\\\text{are
alternating)}}}\\
&  =a_{\left(  6,3,1\right)  }-a_{\left(  5,4,1\right)  }.
\end{align*}
Dividing back by $a_{\rho}$ gives us%
\[
p_{2}s_{\left(  2,2,1\right)  }=s_{\left(  6,3,1\right)  -\rho}-s_{\left(
5,4,1\right)  -\rho}=s_{\left(  4,2,1\right)  }-s_{\left(  3,3,1\right)  }.
\]
\end{example}

So we get an expansion, but the coefficients need not be nonnegative, and we
have to sort our $\alpha+n\overrightarrow{e}_{k}$'s to make them strictly
decreasing again (and remove the ones that have equal entries!). The general
case uses Lemma \ref{lem.schur.alternant-swap} from Lecture 8. If sorting
requires a permutation $\sigma$, then
\[
a_{\alpha+n\overrightarrow{e}_{k}}
=(-1)^{\sigma}a_{(\alpha+n\overrightarrow{e}_{k})\circ\sigma}.
\]
Thus, we keep track of the sign of the sorting permutation. Tuples with
repeated entries contribute zero by part \textbf{(a)} of that lemma.

Back to the combinatorics: can we actually describe the partitions we get in
the expansion? For example, for $N\geq7$, we get%
\begin{align*}
p_{2}s_{\left(  2,2,1\right)  }  &  =s_{\left(  4,2,1\right)  }-s_{\left(
3,3,1\right)  }-s_{\left(  2,2,1,1,1\right)  };\\
p_{3}s_{\left(  3,2,1\right)  }  &  =s_{\left(  3,2,1,1,1,1\right)
}-s_{\left(  3,2,2,2\right)  }-s_{\left(  3,3,3\right)  }-s_{\left(
4,4,1\right)  }+s_{\left(  6,2,1\right)  }.%
\end{align*}
(this still holds when $N<7$, but then some of these Schurs will be $0$). What
is the pattern here? Again, $p_{n}s_{\mu}=\sum\pm s_{\lambda}$ over certain
partitions $\lambda$ for which $\lambda/\mu$ is a skew partition (that is,
$\mu\subseteq\lambda$). But which skew partitions appear in the sum, and with
which signs?

The answer can be described visually in terms of so-called \textbf{ribbons}
aka \textbf{rim hooks} aka \textbf{border strips}. These are skew partitions
\textquotedblleft of width $1$\textquotedblright. There are many equivalent
formal ways to describe them. We will introduce them by giving all these
descriptions and showing that they are equivalent.

\textbf{Notations:}

\begin{itemize}
\item We view the set $\mathbb{Z}^{2}$ as a directed graph with arcs $\left(
i,j\right)  \rightarrow\left(  i+1,j\right)  $ (\textquotedblleft
south-steps\textquotedblright) and $\left(  i,j\right)  \rightarrow\left(
i,j-1\right)  $ (\textquotedblleft west-steps\textquotedblright). This digraph
will be called $\mathbb{Z}_{\operatorname*{sw}}^{2}$. We allow a directed
path to consist of just one vertex (and no arcs).

\item Forgetting the directions of the arcs turns this digraph into an
undirected graph, which we just call $\mathbb{Z}^{2}$.

\item For any $k\in\mathbb{Z}$, the $k$\emph{-th diagonal} means the set of
all $\left(  i,j\right)  \in\mathbb{Z}^{2}$ with $j-i=k$.

\item The \emph{overlap} of two rows $R$ and $S$ of a diagram means the set of
all $j\in\mathbb{Z}$ such that both $R$ and $S$ have a cell in the $j$-th column.

\item If $\lambda$ is a partition, then
\[
R\left(  \lambda\right)  :=\left\{  \left(  i,j\right)  \in Y\left(
\lambda\right)  \ \mid\ \left(  i+1,j+1\right)  \notin Y\left(  \lambda
\right)  \right\}  .
\]
We call this set the \textbf{rim} of $\lambda$. It consists of the cells of
$Y\left(  \lambda\right)  $ that are the
southeasternmost cells on their diagonal to belong to $Y\left(  \lambda
\right)  $.

Moreover, for each $c\in Y\left(  \lambda\right)  $, we set%
\[
R_{c}\left(  \lambda\right)  :=\left\{  \left(  i,j\right)  \in R\left(
\lambda\right)  \ \mid\ \left(  i,j\right)  \geq c\right\}  ,
\]
where the $\geq$ relation refers to the entrywise order on $\mathbb{Z}^{2}$
(that is, \textquotedblleft$\left(  i,j\right)  $ lies weakly southeast of
$c$\textquotedblright).

\item For an integer $N$-tuple $\alpha\in\mathbb{Z}^{N}$, we write
$\operatorname{sort}(\alpha)$ for its entries rearranged in weakly decreasing
order. This is an integer tuple, not necessarily a partition: its entries
may be negative. Repeated entries are not removed.

\item For integers $u,v$, we write $[u,v]:=\{i\in\mathbb{Z}\mid u\leq i\leq v\}$.
In particular, this interval is empty when $u>v$.

\item As usual, $\rho$ means the $N$-partition $\left(  N-1,N-2,\ldots
,1,0\right)  $.
\end{itemize}

\begin{theorem}[Ribbon criteria]\label{thm.schur.ribbon-criteria}
Let $\lambda$ and $\mu$ be two $N$-partitions with $\mu\subseteq\lambda$ and
$n:=\left\vert Y\left(  \lambda/\mu\right)  \right\vert >0$. Then, the
following conditions are equivalent:

\begingroup
\setlength{\leftmargini}{4.8em}
\begin{enumerate}
\item[$\mathcal{R}_{\operatorname*{path}}$:] The diagram $Y\left(  \lambda
/\mu\right)  $ is the set of vertices of a (directed!) path of the digraph
$\mathbb{Z}_{\operatorname*{sw}}^{2}$.

\item[$\mathcal{R}_{\operatorname*{diag}}$:] The diagram $Y\left(  \lambda
/\mu\right)  $ has exactly one cell in each of $n$ consecutive diagonals (and
these exhaust all cells of $Y\left(  \lambda/\mu\right)  $).

\item[$\mathcal{R}_{2\times2}$:] The diagram $Y\left(  \lambda/\mu\right)  $
is connected as a subset of the undirected graph $\mathbb{Z}^{2}$ and contains
no $2\times2$-square $\ydiagram{2,2}$ as a subset.

\item[$\mathcal{R}_{\operatorname*{ovl}}$:] The nonempty rows of the diagram
$Y\left(  \lambda/\mu\right)  $ form an interval (i.e., they are the $u$-th,
$\left(  u+1\right)  $-st, $\ldots$, $v$-th row for some $1\leq u\leq v\leq
N$), and any two consecutive nonempty rows have an overlap of size $1$.

\item[$\mathcal{R}_{+1}$:] There exist $1\leq u\leq v\leq N$ such that
\begin{align*}
\lambda_{i}  &  =\mu_{i-1}+1\ \ \ \ \ \ \ \ \ \ \text{for all }i\in\left[
u+1,v\right]  ,\ \ \ \ \ \ \ \ \ \ \text{and}\\
\lambda_{i}  &  =\mu_{i}\ \ \ \ \ \ \ \ \ \ \text{for all }i\in[N]\setminus\left[
u,v\right]  .
\end{align*}


\item[$\mathcal{R}_{\rho+}$:] We have $\lambda+\rho=\operatorname*{sort}%
\left(  \mu+\rho+n\overrightarrow{e}_{k}\right)  $ for some $k\in\left[
N\right]  $, where $\overrightarrow{e}_{k}\in\mathbb{N}^{N}$ denotes the
vector with a $1$ in its $k$-th position and zeroes everywhere else.

\item[$\mathcal{R}_{\rho-}$:] We have $\mu+\rho=\operatorname*{sort}\left(
\lambda+\rho-n\overrightarrow{e}_{k}\right)  $ for some $k\in\left[  N\right]
$, where $\overrightarrow{e}_{k}\in\mathbb{N}^{N}$ denotes the vector with a
$1$ in its $k$-th position and zeroes everywhere else.

\item[$\mathcal{R}_{R}$:] We have $Y\left(  \lambda/\mu\right)  =R_{c}\left(
\lambda\right)  $ for some $c\in Y\left(  \lambda\right)  $.
\end{enumerate}
\endgroup

Moreover, the $u$ and $v$ in condition $\mathcal{R}_{\operatorname*{ovl}}$ are
precisely the $u$ and $v$ in condition $\mathcal{R}_{+1}$.
\end{theorem}

A skew partition satisfying these equivalent conditions is called a
\textbf{ribbon}, a \textbf{rim hook}, or a \textbf{border strip}. In
particular, a ribbon is nonempty.

\Needspace{13\baselineskip}
\begin{example}
\label{exa.schur.ribbon-criteria}Let $N=4$, $\lambda=(6,5,3,2)$ and
$\mu=(6,2,1,1)$. Thus, $n=6$. The left-hand picture shows the skew diagram,
with its cells numbered in southwest path order. The right-hand picture
shows $Y(\lambda)$, with the rim $R(\lambda)$ shaded and the cell
$c=(2,2)$ marked by a bullet:
\begingroup
\definecolor{ribbonrim}{gray}{0.85}
\[
\begin{array}{c@{\qquad\qquad}c}
Y(\lambda/\mu)&Y(\lambda)\\[6pt]
\begin{ytableau}
\none&\none&\none&\none&\none&\none\\
\none&\none&3&2&1\\
\none&5&4\\
\none&6
\end{ytableau}
&
\begin{ytableau}
{}&{}&{}&{}&*(ribbonrim){}&*(ribbonrim){}\\
{}&\bullet&*(ribbonrim){}&*(ribbonrim){}&*(ribbonrim){}\\
{}&*(ribbonrim){}&*(ribbonrim){}\\
*(ribbonrim){}&*(ribbonrim){}
\end{ytableau}
\end{array}
\]
\endgroup
The path in the left-hand picture is
\[
(2,5)\longrightarrow(2,4)\longrightarrow(2,3)
\longrightarrow(3,3)\longrightarrow(3,2)\longrightarrow(4,2).
\]
Its six cells lie on diagonals $3,2,1,0,-1,-2$, respectively. The diagram is
connected and contains no $2\times2$-square. Its nonempty rows are rows
$2,3,4$, with consecutive overlaps $\{3\}$ and $\{2\}$. Thus, $u=2$ and
$v=4$, and the equalities in $\mathcal{R}_{+1}$ are
\[
\lambda_{3}=3=\mu_{2}+1,\qquad
\lambda_{4}=2=\mu_{3}+1,\qquad \lambda_{1}=\mu_{1}=6.
\]
Also, $Y(\lambda/\mu)=R_{(2,2)}(\lambda)$: we keep precisely the shaded
cells weakly southeast of the bullet. Notice that $c$ itself need not
belong to the rim or to the skew diagram.

Finally, $\rho=(3,2,1,0)$ gives
\[
\mu+\rho=(9,4,2,1),\qquad \lambda+\rho=(9,7,4,2).
\]
Thus, the two sorting conditions hold as well:
\begin{align*}
\operatorname{sort}\left(\mu+\rho+6\overrightarrow{e}_{4}\right)
&=\operatorname{sort}(9,4,2,7)=(9,7,4,2)=\lambda+\rho,\\
\operatorname{sort}\left(\lambda+\rho-6\overrightarrow{e}_{2}\right)
&=\operatorname{sort}(9,1,4,2)=(9,4,2,1)=\mu+\rho.
\end{align*}
The chosen positions in these two conditions need not be the same: here
they are $4$ and $2$. In the first sorting, the $7$ moves left past two
entries, so $a_{(9,4,2,7)}=a_{(9,7,4,2)}$ with a plus sign.
\end{example}

\Needspace{14\baselineskip}
\begin{example}
\label{exa.schur.ribbon-nonexamples}The following two diagrams are not ribbons:
\[
\begin{array}{c@{\qquad\qquad}c}
Y((2,2))&Y((2,1)/(1))\\[5pt]
\ydiagram{2,2}&\ydiagram{1+1,1}\\[6pt]
\begin{array}{c}\text{connected, but contains}\\\text{a }2\times2\text{-square}\end{array}
&\begin{array}{c}\text{no }2\times2\text{-square,}\\\text{but disconnected}\end{array}
\end{array}
\]
The two cells in the second diagram touch only at a corner, which does not
make them adjacent in $\mathbb{Z}^{2}$. They lie on diagonals $1$ and $-1$,
with diagonal $0$ missing. Thus, both clauses in
$\mathcal{R}_{2\times2}$ matter. A diagram with just one cell, on the other
hand, is a ribbon; the empty diagram is not.
\end{example}

\begin{proof}[Beginning of the proof of Theorem \ref{thm.schur.ribbon-criteria}.]
\leavevmode\par\nobreak\noindent
\renewcommand{\qedsymbol}{}
We prove the equivalence of the first three conditions in this lecture.

\smallskip\noindent\textit{Proof of }$\mathcal{R}_{\operatorname*{path}}\Longrightarrow
\mathcal{R}_{\operatorname*{diag}}$\textit{:} Assume $\mathcal{R}%
_{\operatorname*{path}}$. Then, the $n$ cells of $Y\left(  \lambda/\mu\right)
$ are the vertices of a path of $\mathbb{Z}_{\operatorname*{sw}}^{2}$. Each
arc $c\rightarrow d$ of $\mathbb{Z}_{\operatorname*{sw}}^{2}$ has the property
that if $c$ lies on the $k$-th diagonal, then $d$ lies on the $\left(
k-1\right)  $-th diagonal. So the vertices of a path must lie on consecutive
diagonals. In particular, the $n$ cells of $Y\left(  \lambda/\mu\right)  $
must therefore lie on $n$ consecutive diagonals. That is, $\mathcal{R}%
_{\operatorname*{diag}}$ holds.

\smallskip\noindent\textit{Proof of }
$\mathcal{R}_{\operatorname*{diag}}\Longrightarrow
\mathcal{R}_{\operatorname*{path}}$\textit{:} Assume
$\mathcal{R}_{\operatorname*{diag}}$. List the cells as $c_{1},\ldots,c_{n}$
so that $c_{i}$ lies on diagonal $d+i$, for some integer $d$. We shall show
that there is an arc from $c_{i}$ to $c_{i-1}$ for each $i\in\{2,\ldots,n\}$.
Then $c_{n},c_{n-1},\ldots,c_{1}$ is the required path in
$\mathbb{Z}_{\operatorname*{sw}}^{2}$.

Fix $i\in\{2,\ldots,n\}$ and write $c_{i}=(a,b)$. Since $c_{i-1}$ lies on
the preceding diagonal, it has the form
\[
c_{i-1}=(a+t,b+t-1)\qquad\text{for some }t\in\mathbb{Z}.
\]
Let $W=(a,b-1)$ and $S=(a+1,b)$ be the western and southern neighbors of
$c_{i}$. The possibilities $t=0$ and $t=1$ give $c_{i-1}=W$ and
$c_{i-1}=S$, respectively, and therefore give the desired arc. All other
positions lie farther northwest or southeast on the same diagonal:
\[
\begin{tikzpicture}[x=1cm,y=1cm,
  cell/.style={draw,fill=white,minimum size=0.48cm,inner sep=1pt,font=\small}]
\draw[densely dotted] (-2.5,1.5)--(1.5,-2.5);
\node[cell] (nw) at (-2,1) {$?$};
\node[anchor=east,font=\small] at (-2.35,1) {$t<0$};
\node[cell] (w) at (-1,0) {$W$};
\node[anchor=east,font=\small] at (-1.35,0) {$t=0$};
\node[cell] (s) at (0,-1) {$S$};
\node[anchor=west,font=\small] at (0.35,-1) {$t=1$};
\node[cell] (se) at (1,-2) {$?$};
\node[anchor=west,font=\small] at (1.35,-2) {$t>1$};
\node[cell] (c) at (0,0) {$c_i$};
\draw[-{Stealth[length=4pt]}] (c.west)--(w.east);
\draw[-{Stealth[length=4pt]}] (c.south)--(s.north);
\end{tikzpicture}
\]
The two question marks illustrate positions with $t<0$ and $t>1$; such
positions can be arbitrarily far along the dotted diagonal.

If $t<0$, then $c_{i-1}\leq W\leq c_i$ in the northwest order. By the
convexity of skew diagrams (Lemma \ref{lem.schur.skew-convexity}), the cell
$W$ must also belong to $Y(\lambda/\mu)$. But $W$ and $c_{i-1}$ are
distinct cells on the same diagonal, contradicting
$\mathcal{R}_{\operatorname*{diag}}$.

If $t>1$, then $c_i\leq S\leq c_{i-1}$. Convexity now puts $S$ in the
diagram, again giving two distinct cells, $S$ and $c_{i-1}$, on the same
diagonal. This is the same contradiction. Therefore $t\in\{0,1\}$, as
required. For $n=1$, there are no arcs to check and the one cell is already
a path.

Thus, $\mathcal{R}_{\operatorname*{path}}\Longleftrightarrow\mathcal{R}%
_{\operatorname*{diag}}$.

\smallskip\noindent\textit{Proof of }$\mathcal{R}_{\operatorname*{diag}}\Longrightarrow
\mathcal{R}_{2\times2}$\textit{:} Assume $\mathcal{R}_{\operatorname*{diag}}$
holds. Then, there are no $2\times2$-squares in $Y\left(  \lambda/\mu\right)
$ because a $2\times2$-square would contain two cells on the same diagonal.
Moreover, $\mathcal{R}_{\operatorname*{diag}}\Longrightarrow\mathcal{R}%
_{\operatorname*{path}}$ shows that we have $\mathcal{R}_{\operatorname*{path}%
}$, and therefore $Y\left(  \lambda/\mu\right)  $ is connected (since a path
is connected). So $\mathcal{R}_{2\times2}$ holds.

\smallskip\noindent\textit{Proof of }
$\mathcal{R}_{2\times2}\Longrightarrow
\mathcal{R}_{\operatorname*{diag}}$\textit{:} Assume
$\mathcal{R}_{2\times2}$. Adjacent cells of $\mathbb{Z}^{2}$ lie on
adjacent diagonals. Since $Y(\lambda/\mu)$ is connected, any two of its
cells can be joined by a path within the diagram. Along such a path the
diagonal number changes by $1$ or $-1$ at each step, so it takes every
integer value between its starting and ending values. Consequently, the
diagonals met by $Y(\lambda/\mu)$ form an interval of integers.

Moreover, no two cells of $Y(\lambda/\mu)$ can lie on the same diagonal.
Indeed, two distinct cells on the same diagonal have the form $(a,b)$ and
$(a+t,b+t)$ for some integer $t\geq1$, after interchanging them if needed.
Every cell of the square
\[
\{(a,b),(a,b+1),(a+1,b),(a+1,b+1)\}
\]
lies between these two cells in the northwest order. Convexity would
therefore put this entire $2\times2$-square in $Y(\lambda/\mu)$, contrary
to our assumption. Thus, there is exactly one cell on each occupied
diagonal; since there are $n$ cells, there are exactly $n$ consecutive
occupied diagonals. This proves $\mathcal{R}_{\operatorname*{diag}}$.

\smallskip
\textbf{Please try to think about the other equivalences.} We will finish
this proof next time and then state the Murnaghan--Nakayama rule for
$p_{n}s_{\mu}$ in terms of ribbons.
\end{proof}


\end{document}