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[Torus orbit closures and 1-strip-less-tableaux]
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{Torus orbit closures and 1-strip-less-tableaux}
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\author[\initial{C.} Lian]{\firstname{Carl} \lastname{Lian}}
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\address{Tufts University\\
Department of Mathematics\\
177 College Ave\\
Medford, MA 02155 (USA)}
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%\curraddr{Coll\`ege Royal Henri-Le-Grand\\
%Prytan\'ee National Militaire\\
%72000 La Fl\`eche, Sarthe (France)}
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\email{Carl.Lian@tufts.edu}
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\urladdr{https://sites.google.com/view/carllian}
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%\author{\firstname{Joseph} \lastname{Fourier}}
%\address{Universit\'e de Grenoble\\
% Institut Moi-m\^eme\\
% BP74, 38402 SMH Cedex (France)}
%\email{fourier@fourier.edu.fr}
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\keywords{torus orbit, Grassmannian, Littlewood-Richardson, Young tableau}
%% Mathematical classification (2010)
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\subjclass{10X99, 14A12, 11L05}
\datepublished{2024-09-03}
\begin{document}
%% Abstracts must be placed before \maketitle
\begin{abstract}
We compare two formulas for the class of a generic torus orbit closure in a Grassmannian, due to Klyachko and Berget-Fink. The naturally emerging combinatorial objects are semi-standard fillings we call \emph{1-strip-less tableaux}.
\end{abstract}
\maketitle
% First paragraph after a section is not indented. If there is text
% below the title before the first section, it should be unindented
% like this.
\section{Introduction}
Let $\Gr(r,n)$ be the Grassmannian of $r$-planes in $\bC^n$, on which an $n$-dimensional torus $T$ acts in the standard way. Let $x\in \Gr(r,n)$ be a general point, let $X:=\overline{T\cdot x}$ be the closure of the orbit of $x$, and let $[X]\in H^{2(r-1)(n-r-1)}(\Gr(r,n))$ be the corresponding cohomology class. Such classes have received considerable attention in the literature \cite{K85,Spe09,BF17,LPST20,NT23}; the author's own interest in them stems from their relationship to fixed-domain curve counts in the projective space $\bP^{r-1}$ \cite{L23}.
We have the following two formulas for $[X]$ originally due to Klyachko and Berget-Fink, respectively, obtained via different methods.
\begin{thm}[{\cite[Theorem 6]{K85}}]\label{k_formula}
Write
\begin{equation*}
[X]=\sum_{\mu} \Gamma^\mu_{r,n}\sigma_{\mu}
\end{equation*}
in terms of the basis of Schubert cycles $\sigma_\mu$. Then, we have
\begin{equation*}
\Gamma^\mu_{r,n}=\sum_{j=0}^{m(\mu)}(-1)^j\binom{n}{j}|\ssyt_{r-j}(\mu^j)|,
\end{equation*}
where:
\begin{itemize}
\item $m(\mu)$ denotes the number of parts of $\mu$ equal to $n-r$ (the largest possible),
\item $\mu^j$ denotes the partition of length $r-j$ obtained from $\mu$ by removing $j\le m(\mu)$ parts of size $n-r$, and
\item $|\ssyt_{r'}(\mu')|$ denotes the number of semi-standard Young tableaux (SSYTs) of shape $\mu'$ filled with entries $1,2,\ldots,r'$.
\end{itemize}
\end{thm}
See \S\ref{conventions} for a detailed discussion of notational conventions for Schubert calculus and related combinatorics.
\begin{thm}[{\cite[Theorem 5.1]{BF17}}]\label{bf_formula}
We have
\begin{equation*}
[X]=\sum_{\lambda\subset(n-r-1)^{r-1}}\sigma_{\lambda}\sigma_{\widetilde{\lambda}},
\end{equation*}
where, for any partition $\lambda\subset(n-r-1)^{r-1}$, $\widetilde{\lambda}$ denotes its complement inside the rectangle $(n-r-1)^{r-1}$.
\end{thm}
Therefore, for abstract reasons, the following holds.
\begin{thm}\label{main_formula}
We have
\begin{equation*}
\sum_{\lambda\subset(n-r-1)^{r-1}}\sigma_{\lambda}\sigma_{\widetilde{\lambda}}=\sum_{\mu} \Gamma^\mu_{r,n}\sigma_{\mu},
\end{equation*}
where the $\Gamma^\mu_{r,n}$ are as in Theorem \ref{k_formula}.
\end{thm}
The purpose of this paper is to give a direct combinatorial proof of Theorem \ref{main_formula}. Our main calculation applies Coskun's geometric Littlewood-Richardson rule \cite{Cos09} to the left-hand side in such a way that the coefficients $\Gamma^\mu_{r,n}$ emerge naturally. In fact, we arrive at the following combinatorial interpretation for the coefficients $\Gamma^\mu_{r,n}$:
\begin{thm}\label{stripless_interpretation}
$\Gamma^\mu_{r,n}$ is equal to the number of \textit{1-strip-less} SSYTs (see \S\ref{yt_section}) of shape $\mu$, filled with entries $1,2,\ldots,r$.
\end{thm}
In particular, in the alternating sum of Theorem \ref{k_formula}, the first term dominates, and we get a \emph{non-negative} interpretation of the integers $\Gamma^\mu_{r,n}$. In fact, the following version of Theorem \ref{stripless_interpretation} was known to Klyachko \cite[Theorem 3.3]{K95}: the number $\Gamma^\mu_{r,n}$ was shown to count the number of \emph{standard} Young tableaux of shape $\overline{\lambda}$, the complement of $\lambda$ inside the rectangle $(n-r)^r$, with exactly $(r-1)$ descents. Such objects are naturally in bijection with 1-strip-less SSYTs; we describe this bijection (explained to us by Philippe Nadeau) in the appendix.
In our calculation, the 1-strip-less SSYTs are the naturally occurring objects. As we will see in \S\ref{LR_sec}, the geometric Littlewood-Richardson rule allows us to express the products $\sigma_{\lambda}\sigma_{\widetilde{\lambda}}$ in terms of simple Schubert classes, see Corollary \ref{explicit_M}. From here, the Pieri rule allows us on the one hand to interpret the coefficients in the Schubert basis as counts of 1-strip-less tableaux in \S\ref{stripless_sec}, and on the other hand to recover Klyachko's formula in \S\ref{klyachko_comparison_sec}.
\section{Preliminaries}\label{conventions}
\subsection{Young tableaux}\label{yt_section}
Let $\lambda=(\lambda_1,\ldots,\lambda_r)$ be a partition. Our convention throughout is that the parts of $\lambda$ are non-increasing $(\lambda_1\ge\cdots\ge\lambda_r\ge0)$, and furthermore that $\lambda$ is defined implicitly with respect to an integer $n$ for which $\lambda_1\le n-r$ (equivalently, $\lambda\subset(n-r)^r$). The Young diagram of $\lambda$ is taken to be left- and upward-justified, with larger parts depicted on top.
The \emph{complement} of $\lambda$, denoted $\overline{\lambda}$, is the partition $(n-r-\lambda_r,\ldots,n-r-\lambda_1)$. Pictorially, the complement $\overline{\lambda}$ is obtained as the (rotated) complement of $\lambda$ inside the rectangle $(n-r)^r$. Note that we distinguish the notation $\overline{\lambda}$ and the notation $\widetilde{\lambda}$ appearing in Theorem \ref{bf_formula}.
Below, the Young diagram of the partition $\lambda=(10,9,4,2)$, defined with respect to $n=14$ is shown. Its complement (before rotation) $\overline{\lambda}=(8,6,1,0)$ is shaded in gray.
\begin{equation*}
\ydiagram[*(white) ]
{10,9,4,2}
*[*(gray)]{10,10,10,10}
\end{equation*}
A \emph{strip} $S$ of $\lambda$ is a subset of $n-r$ boxes in the Young diagram of $\lambda$ satisfying the following two properties:
\begin{itemize}
\item No two boxes of $S$ lie in the same column.
\item Given any distinct boxes $b_1,b_2$ of $S$, if $b_1$ lies in a column to the left of $b_2$, then $b_1$ does not also lie in a row above $b_2$.
\end{itemize}
An example of a strip in the partition $\lambda=(10,9,4,2)$, where we take $n=14$, is shaded below. However, the same set of boxes is \emph{not} a strip if $\lambda$ is defined with respect to $n>14$.
%An equivalent characterization of a partial strip is a subset boxes in the Young diagram of $\lambda$ for which there exists a semi-standard Young tableau (defined below) of shape $\lambda$ and an integer $j$, such that the boxes filled with the integer $j$ are exactly those in $S$. An example of a partial strip in the partition $\lambda=(10,9,4,2)$ is shaded below.
%\begin{equation*}
%\ydiagram[*(white) ]
%{6,4}
%*[*(gray)]{6+3,4+2,1+2}
%*[*(white)]{9+0+1,6+3,4,2}
%\end{equation*}
%
%A \emph{strip} $S$ of $\lambda$ is a partial strip that has $n-r$ boxes, the maximum possible. In this case, property (S3) is immediate from properties (S1) and (S2). An example of a strip in the partition $\lambda=(10,9,4,2)$, where we take $n=14$, is shaded below. However, the same set of boxes is \emph{not} a strip if $\lambda$ is defined with respect to $n>14$, but only a partial strip.
\begin{equation*}
\ydiagram[*(white) ]
{6,4}
*[*(gray)]{6+4,4+2,4}
*[*(white)]{10+0,6+3,4+0,2}
\end{equation*}
%We will deal primarily with strips, but partial strips will arise in Proposition \ref{term_stripless} in the context of the Pieri rule.
A \emph{semi-standard Young tableau (SSYT)} of shape $\lambda$ is a filling of the boxes of $\lambda$ with the entries $1,2,\ldots,r$ so that entries increase weakly across rows and strictly down columns. We will often abuse notation, using the letter $\lambda$ to denote either a partition or a SSYT of that shape. The number of SSYTs of shape $\lambda$ is denoted $|\ssyt_r(\lambda)|$, and is given by the formula
\begin{equation*}
|\ssyt_r(\lambda)|=s_\lambda(1^r)=\prod_{u\in\lambda}\frac{r+c(u)}{h(u)},
\end{equation*}
see \cite[Corollary 7.21.4]{EC2}. Here, $s_\lambda(1^r)$ denotes the Schur function associated to $\lambda$, specialized so that the first $r$ variables are equal to 1 and all others are equal to zero. In the last formula, the product is over all boxes $u$, and if $u$ is in the $i$-th row and $j$-th column of $\lambda$, then by definition, $c(u)=j-i$ and $h(u)$ is the total number of boxes either to the right (and in the same row) of, below (and in the same column), or equal to $u$.
The \emph{$k$-weight} $w_k(\lambda)$ of a SSYT $\lambda$ is the number of appearances of the entry $k$. The \emph{type} of $\lambda$ is the tuple $(w_1(\lambda),w_2(\lambda),\ldots,w_r(\lambda))$.
For $i=1,2,\ldots,r$, an \emph{$(i)$-strip} of a SSYT is a strip, all of whose boxes are filled with the entry $i$. A \emph{0-strip} (written without parentheses) is, by definition, an $(i)$-strip for some $i$. A SSYT is \emph{0-strip-less} if it contains no 0-strips. Below, the SSYT of shape $\lambda=(10,9,4,2)$ (with $n=14$) has a $(3)$-strip and no other 0-strips. If we take instead $n>14$, then the same SYT is 0-strip-less.
\begin{equation*}
\begin{ytableau}
1 & 1 & 1 & 1 & 1 & 2 & 3 & 3 & 3 & 3 \\
2 & 2 & 2 & 2 & 3 & 3 & 4 & 4 & 4\\
3 & 3 & 3 & 3\\
4 & 4
\end{ytableau}
\end{equation*}
Let $|\ssyt^0_r(\lambda)|$ denote the number of 0-strip-less SSYTs of shape $\lambda$ (with entries $1,2,\ldots,r$).
\begin{lemma}\label{count_0strip}
Given a partition $\lambda\subset(n-r)^r$, adopt the notation $m(\lambda)$ and $\lambda^j$, for $j=0,1,\ldots,m(\lambda)$, of the statement of Theorem \ref{k_formula}. Then, we have
\begin{equation*}
|\ssyt^0_r(\lambda)|=\sum_{j=0}^{m(\lambda)}(-1)^j\binom{r}{j}|\ssyt_{r-j}(\lambda^j)|.
\end{equation*}
\end{lemma}
\begin{proof}
Given any $j$-element subset of $S\subset \{1,2,\ldots,r\}$, the number $|\ssyt_{r-j}(\mu^j)|$ counts SSYTs of shape $\mu$ with an $(s)$-strip for any $s\in S$. Indeed, deleting all boxes containing an entry $s\in S$ in such a SSYT yields (after appropriate re-shifting) an SSYT of shape $\mu^j$ filled with entries lying in $\{1,2,\ldots,r\}-S$; this is easily seen to be a bijection. The factor $\binom{r}{j}$ enumerates subsets $S$ of size $j$. The lemma now follows from Inclusion-Exclusion.
\end{proof}
Similarly, for $i=1,2,\ldots,r-1$, an \emph{$(i,i+1)$-strip} of a SSYT is a strip, all of whose boxes are filled with the entry $i$ or $i+1$, and for which all instances of $i$ all appear to the left of all instances of $i+1$. By convention, an $(i)$-strip is both an $(i,i+1)$- and an $(i-1,i)$-strip. A \emph{1-strip} (written without parentheses) is, by definition, an $(i,i+1)$-strip for some $i$. A SSYT is \emph{1-strip-less} if it contains no 1-strips (and therefore no 0-strips). Note that an SSYT of shape $\lambda$ with $\lambda_11$, but with opposite signs ($r$ differs by 1, but $\ell$ does not change). Indeed, the extra factor of $\sigma_{n-1}$ is absorbed into the $j=1$ factor in each summand. This completes the proof.
\end{proof}
%Coskun's Littlewood-Richardson Rule performs a sequence of modifications to $M$ until it is a formal sum of \emph{nested} Mondrian tableau corresponding to Schubert cycles. Rather than carrying this out, we compare $M$ to the Mondrian tableau $M'$ with \emph{disjoint} squares $M'_j$ containing the basis elements $e_{a_{j-1}-(r-j)},\ldots,e_{a_j-(r-j)}$, depicted in Figure \ref{mondrian3}. The square $M'_j$ is obtained from $M_j$ by shifting by $r-j$ units to the southwest.
\begin{rema}\label{subvarieties_break_rmk}
We explain the geometric content of the formula \eqref{subvarieties_break}. The claim amounts to studying the degeneration of the subvariety $Z_\circ\subset \Gr(r,n+r-1)$ under the degeneration of the basis $e_{-(r-2)},\ldots,e_n$ that replaces $e_0$ with $e_0^t=te_0+(1-t)e_{a_1}$ at time $t$. When $t\neq0$, one has a closed subvariety $Z^t_\circ\subset \Gr(r,n+r-1)$ defined in the same way as $Z_\circ$, now with respect to the basis $e_{-(r-2)},\ldots,e_0^t,\ldots,e_n$. When $t=0$, these vectors are no longer linearly independent, but we may still study the limit $Z^{\lim}_\circ:=\lim_{t\to 0}Z^t_\circ$.
In the limit, the analysis in the proof of \cite[Theorem 3.32]{Cos09} shows that one of two things may happen generically. The first possibility is that the limits of the vectors $v_1,\ldots,v_r$ remain linearly independent, and we simply have the new condition that $v_1$ lie in the new subspace $\langle e_1,\ldots,e_{a_1}\rangle$, equal to the limit of the subspace $\langle e_0,\ldots,e_{a_1-1}\rangle$ as $t\to 0$. In this way, $Z$ appears as a component of the limit of $Z^{\lim}_\circ$. Second, it may happen that the vectors $v_1$ and $v_2$ become linearly dependent, necessarily both constant multiples of the basis vector $e_{a_1}$, which spans the intersection $\langle e_1,\ldots,e_{a_1}\rangle\cap \langle e_{a_1},\ldots,e_{a_2}\rangle$. If this is the case, then the limit of the span $\langle v_1,v_2\rangle$ must also continue to lie in $\langle e_0,\ldots,e_{a_2}\rangle$, as this holds for every $t\neq0$. This exhibits $Z_+$ as the other component of the limit of $Z^{\lim}_\circ$. Both components are shown in \cite{Cos09} to appear with multiplicity 1, and one concludes that $[Z_\circ]=[Z^{\lim}_\circ]=[Z]+[Z_+]$.
\end{rema}
\section{1-strip-less tableaux}\label{stripless_sec}
We now consider the individual terms appearing on the right hand side of Corollary \ref{explicit_M}.
\begin{prop}\label{term_stripless}
Consider the expansion of
\begin{equation*}
\prod_{j=1}^{\ell}\sigma_{(n-1)^{s_j-s_{j-1}-1},n-1-(a_{s_j}-a_{s_{j-1}})}
\end{equation*}
in $H^{*}(\Gr(r,n+r-1))$ as a linear combination of Schubert cycles $\sigma_\mu$, with $\mu\subset(n-1)^{r}$ and $|\mu|=(n-1)(r-1)$.
Then, the coefficient of $\sigma_\mu$ is equal to the number of SSYTs of shape $\mu$ with the properties that:
\begin{itemize}
\item the $i$-weight $w_i(\mu)$ (that is, the number of boxes of $\mu$ filled with the entry $i$) is equal to $n-1-(a_{i}-a_{i-1})$, for $i=1,2,\ldots,r$, and
\item $\mu$ contains an $(i,i+1)$-strip, for all $i\in\{1,2,\ldots,r-1\}-\{s_1,\ldots,s_{\ell-1}\}$.
\end{itemize}
\end{prop}
Before giving the proof, we first give an illustrative example. Take $n=18$, $r=6$, and $(a_0,a_1,a_2,a_3,a_4,a_5,a_6)=(1,3,7,8,10,14,18)$. Take also $\ell=3$ and $(s_0,s_1,s_2,s_3)=(0,2,5,6)$. Then, the class in question is
\begin{equation*}
\sigma_{17,11}\cdot\sigma_{17,17,10}\cdot\sigma_{13}\in H^{2\cdot5\cdot17}(\Gr(6,23)),
\end{equation*}
which we re-write as
\begin{equation*}
\sigma_{17}\cdot\sigma_{11}\cdot\sigma_{17}\cdot\sigma_{17}\cdot\sigma_{10}\cdot\sigma_{13}.
\end{equation*}
By the Pieri rule, the coefficient of $\sigma_\mu$ in the expansion of this class is equal to the number of SSYTs of shape $\mu$ and type $(17,11,17,17,10,13)$. One such, for $\mu=(17,17,17,16,13,5)$, is shown below:
\begin{equation*}
\begin{ytableau}
1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 \\
2 & 2 & 2 & 2 & 2 & 2 & 2 & 2 & 2 & 2 & 2 & 3 & 3 & 3 & 3 & 3 & 3 \\
3 & 3 & 3 & 3 & 3 & 3 & 3 & 3 & 3 & 3 & 3 & 4 & 4 & 4 & 4 & 4 & 4 \\
4 & 4 & 4 & 4 & 4 & 4 & 4 & 4 & 4 & 4 & 4 & 5 & 5 & 5 & 5 & 6\\
5 & 5 & 5 & 5 & 5 & 5 & 6 & 6 & 6 & 6 & 6 & 6 & 6\\
6 & 6 & 6 & 6 & 6
\end{ytableau}
\end{equation*}
Let $\mu^i$ denote the partition obtained by restricting to the boxes filled with entries that are most $i$. Note that, if $w_i(\mu)=n-1=17$, then the partition $\mu^{i-1}$ determines the 17 boxes filled with the entry $i$. Thus, the sequence of partitions $\mu^2\subset\mu^5\subset\mu^6=\mu$ completely determines the filling of $\mu$ with type $(17,11,17,17,10,13)$.
We now describe a re-filling of the above shape in such a way that $w_i(\mu)=n-1-(a_{i}-a_{i-1})$ for $i=1,\ldots,6$, that is, $\mu$ instead has type $(15,13,16,15,13,13)$, in such a way that preserves the partitions $\mu^2\subset\mu^5\subset\mu^6=\mu$.
First, note that the shape $\mu^2=(17,11)$ consists of $n-1-(a_{s_1}-a_{s_0})=11$ columns with $s_1-s_0=2$ boxes and the remaining six columns have one box. The 11 columns with two boxes must be filled downward with the entries $1,2$. Of the remaining six columns, four must contain the entry 1 (or equivalently, be missing the entry 2) and two must contain the entry 2 (be missing the entry 1). We fill these six columns from left to right, first the four missing the entry 2, then the remaining two missing the entry 1. The resulting filling of $\mu^2$ is
\begin{equation*}
\begin{ytableau}
1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 2 & 2 \\
2 & 2 & 2 & 2 & 2 & 2 & 2 & 2 & 2 & 2 & 2
\end{ytableau}
\end{equation*}
We next fill the skew shape $\mu^5/\mu^2=(17,17,17,15,6)/(17,11)$. It consists of $n-1-(a_{s_2}-a_{s_1})=10$ columns with $s_2-s_0=3$ boxes and seven remaining columns have two boxes each. Those with three boxes must be filled downward with the entries $3,4,5$. Of the remaining seven columns, four must be missing the entry 5, two must be missing the entry 4, and one must be missing the entry 3. We place these columns this order from left to right, and fill each in the unique possible way, that is, with entries increasing down columns. The resulting filling of $\mu^5$ (when combined with that of $\mu^2$ is:
\begin{equation*}
\begin{ytableau}
1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 2 & 2 \\
2 & 2 & 2 & 2 & 2 & 2 & 2 & 2 & 2 & 2 & 2 & 3 & 3 & 3 & 3 & \textbf{3} & \textbf{4} \\
3 & 3 & 3 & 3 & 3 & 3 & \textbf{3} & \textbf{3} & \textbf{3} & \textbf{3} & \textbf{3} & 4 & 4 & 4 & 4 & \textbf{5} & \textbf{5} \\
4 & 4 & 4 & 4 & 4 & 4 & \textbf{4} & \textbf{4} & \textbf{4} & \textbf{4} & \textbf{5} & 5 & 5 & 5 & 5\\
5 & 5 & 5 & 5 & 5 & 5
\end{ytableau}
\end{equation*}
The entries of the columns of $\mu^5/\mu^2$ containing only two entries are shown in bold.
Finally, the skew shape $\mu^6/\mu^5$ is filled in the only way possible, with the entry 6 appearing in all 13 boxes:
\begin{equation*}
\begin{ytableau}
1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 2 & 2 \\
2 & 2 & 2 & 2 & 2 & 2 & 2 & 2 & 2 & 2 & 2 & 3 & 3 & 3 & 3 & \textbf{3} & \textbf{4} \\
3 & 3 & 3 & 3 & 3 & 3 & \textbf{3} & \textbf{3} & \textbf{3} & \textbf{3} & \textbf{3} & 4 & 4 & 4 & 4 & \textbf{5} & \textbf{5} \\
4 & 4 & 4 & 4 & 4 & 4 & \textbf{4} & \textbf{4} & \textbf{4} & \textbf{4} & \textbf{5} & 5 & 5 & 5 & 5\\
5 & 5 & 5 & 5 & 5 & 5 & 6 & 6 & 6 & 6 & 6 & 6 & 6\\
6 & 6 & 6 & 6 & 6
\end{ytableau}
\end{equation*}
We observe in addition that this new SSYT $\mu$ has an $(i,i+1)$-strip for $i=1,3,4$. The $(1,2)$-strip is clearly visible in the top row; we describe how to find $(3,4)$- and $(4,5)$-strips, necessarily inside $\mu^5/\mu^2$. Note that there are two contiguous blocks of columns of $\mu^5/\mu^2$ of size 3. Consider first the leftmost six columns, which are bordered on the right by a column of $\mu^5/\mu^2$ missing the entry 5. Then, we make bold the entries other than 5 in the leftmost six columns. Similarly, in columns 12 through 15, which are bordered on both sides by columns of $\mu^5/\mu^2$ missing the entry 4, we make bold the entries other than 4. We obtain now
\begin{equation*}
\begin{ytableau}
1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 2 & 2 \\
2 & 2 & 2 & 2 & 2 & 2 & 2 & 2 & 2 & 2 & 2 & \textbf{3} & \textbf{3} & \textbf{3} & \textbf{3} & \textbf{3} & \textbf{4} \\
\textbf{3} & \textbf{3} & \textbf{3} & \textbf{3} & \textbf{3} & \textbf{3} & \textbf{3} & \textbf{3} & \textbf{3} & \textbf{3} & \textbf{3} & 4 & 4 & 4 & 4 & \textbf{5} & \textbf{5} \\
\textbf{4} & \textbf{4} & \textbf{4} & \textbf{4}& \textbf{4} & \textbf{4} & \textbf{4} & \textbf{4} & \textbf{4} & \textbf{4} & \textbf{5} & \textbf{5} & \textbf{5} & \textbf{5} & \textbf{5} \\
5 & 5 & 5 & 5 & 5 & 5 & 6 & 6 & 6 & 6 & 6 & 6 & 6\\
6 & 6 & 6 & 6 & 6
\end{ytableau}
\end{equation*}
Now, each column of $\mu^5/\mu^2$ has exactly two entries in bold. The upper bold entries in each column form a $(3,4)$-strip, and the lower bold entries in each column form a $(4,5)$-strip.
The proof below describes an analogous algorithm in general, producing for each term of the product
\begin{equation*}
\prod_{j=1}^{\ell}\sigma_{(n-1)^{s_j-s_{j-1}-1},n-1-(a_{s_j}-a_{s_{j-1}})}
\end{equation*}
a SSYT of shape $\mu$ with the desired type and strips. We will show in addition that this algorithm gives a bijection between the terms in the product at hand and the needed SSYTs.
\begin{proof}[Proof of Proposition \ref{term_stripless}]
The Pieri rule implies that the terms $\sigma_\mu$ in the expansion of
\begin{equation*}
\prod_{j=1}^{\ell}\sigma_{(n-1)^{s_j-s_{j-1}-1},n-1-(a_{s_j}-a_{s_{j-1}})}
\end{equation*}
are in bijection with SSYTs of shape $\mu$ with $w_{s_j}(\mu)=n-1-(a_{s_j}-a_{s_{j-1}})$ for $j=1,\ldots,\ell$ and $w_{i}(\mu)=n-1$ for all other values of $i\in\{1,\ldots,r\}$.
Such an SSYT $\mu$ is determined by the data of the sequence of subpartitions
\begin{equation*}
\mu^{s_1}\subset\mu^{s_2}\cdots\subset\mu^{s_\ell}=\mu,
\end{equation*}
where $\mu^i$ is, as before, the collection of boxes of $\mu$ with entry at most $i$. Furthermore, the skew shape $\mu^{s_j}/\mu^{s_{j-1}}$ has the property that $a_{s_j}-a_{s_{j-1}}$ of its columns have size $s_j-s_{j-1}-1$, while the remaining $n-1-(a_{s_j}-a_{s_{j-1}})$ have size $s_j-s_{j-1}$. We will refer to the former columns as \emph{short} and the latter as \emph{tall}.
We now describe a filling of the same shape $\mu$, by re-filling each skew-shape $\mu^{s_j}/\mu^{s_{j-1}}$ with the entries $s_{j-1}+1,\ldots,s_j$. The tall columns are filled in the only semi-standard way possible, with the entries $s_{j-1}+1,\ldots,s_j$ from top to bottom. The short columns are filled such that the unique missing entries of each column are non-increasing from left to right, and the entries increase going down each column. The short columns may furthermore be filled in a unique way such that the weight of $i$ is equal to $n-1-(a_{i}-a_{i-1})$ for all $i\in[s_{j-1}+1,s_j]$, because the total number of squares of $\mu$ is
\begin{equation*}
\sum_{i=s_{j-1}+1}^{s_j}[n-1-(a_{i}-a_{i-1})]=(n-1)(s_j-s_{j-1}-1)+[(n-1)-(a_{s_j}-a_{s_{j-1}})].
\end{equation*}
We claim that the new filling is a SSYT; it suffices to restrict our attention to the individual skew-shapes $\mu^{s_j}/\mu^{s_{j-1}}$. The entries are increasing down columns by construction. Between consecutive columns of the same height (tall or short), the entries are weakly increasing across rows if the two columns are flush with each other, and this remains true if the column on the right is shifted upward. If a tall column lies immediately to the left of a short column, the entries are weakly increasing across rows because the highest box of the short column cannot lie below that of the tall column. Similarly, if a short column lies immediately to the left of a tall column, we get the same conclusion because the lowest box of the short column cannot lie above that of the short column.
We next show that the new filling contains the needed 1-strips, by making some of the entries of $\mu^{s_j}/\mu^{s_{j-1}}$ bold. First, all entries in each short column are made bold. Then, for each contiguous block of tall columns (not necessarily all flush with each other), suppose that the short columns to the immediate left and right are missing the entries $p\ge q$, respectively. We take $p=s_j$ if our block of tall columns includes the leftmost column of $\mu$, and $q=s_{j-1}$ if it includes the rightmost column of $\mu$. Now, pick any integer $s\in[q,p]$, and make all entries of the block of the tall columns bold except $s$. In this way, each column of $\mu^{s_j}/\mu^{s_{j-1}}$ contains exactly $s_j-s_{j-1}-1$ bold entries.
We claim that, for $t=1,2,\ldots,s_j-s_{j-1}-1$, taking the $t$-th bold entry in each column of $\mu^{s_j}/\mu^{s_{j-1}}$ gives a $(s_{j-1}+t,s_{j-1}+t+1)$-strip. By construction, these entries must each be equal to one of $s_{j-1}+t,s_{j-1}+t+1$. More precisely, if the missing (in a short column) or unique non-bold (in a tall column) entry is strictly greater than $s_{j-1}+t$, then the $t$-th bold entry is equal to $s_{j-1}+t$, and the $t$-th bold entry is otherwise equal to $s_{j-1}+t+1$. Because the missing and non-bold entries do not increase from left to right, this sequence of bold entries is given by some number of instances of $s_{j-1}+t$, followed by the remaining number of instances of $s_{j-1}+t+1$. (Note that both numbers must be non-zero, because $w_i(\mu)=n-1-(a_i-a_{i-1})