%% if you are submitting an initial manuscript then you should have submission as an option here
%% if you are submitting a revised manuscript then you should have revision as an option here
%% otherwise options taken by the article class will be accepted
\documentclass[finalversion]{FPSAC2025}
\articlenumber{102}
%% but DO NOT pass any options (or change anything else anywhere) which alters page size / layout / font size etc

%% note that the class file already loads {amsmath, amsthm, amssymb}


\newtheorem{theorem}{Theorem}[section]
\newtheorem{theorems}[theorem]{Theorems}
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{proposition}[theorem]{Proposition}
\newtheorem{corollary}[theorem]{Corollary}
\newtheorem{example}[theorem]{Example}
\newtheorem{remark}[theorem]{Remark}

\usepackage{lipsum}


\usepackage{amsmath, amsthm, latexsym, amssymb, graphicx, verbatim, mathrsfs,  tikz-cd, bbm, upgreek, accents}
%\usepackage{hyperref}
%\hypersetup{colorlinks=true}
%\usepackage{todonotes}
\usepackage{stackengine}
\usepackage{scalerel}
\usepackage{standalone}

\usepackage{ytableau}
\usepackage{youngtab}

\usepackage{wrapfig}



\usepackage{caption, subcaption}
%\captionsetup{format=plain, font=normalsize, labelfont=bf}
%\captionsetup{width=1.1\linewidth}




%Helpful commands
\newcommand{\R}{\mathbb{R}}
\newcommand{\N}{\mathbb{N}}
\newcommand{\C}{\mathbb{C}}
\newcommand{\Q}{\mathbb{Q}}
\newcommand{\Z}{\mathbb{Z}}
\newcommand{\F}{\mathbb{F}}
\newcommand{\PP}{\mathbb{P}}
\newcommand{\ra}{\rightarrow}

\newcommand{\Acal}{\mathcal{A}}
\newcommand{\Bcal}{\mathcal{B}}
\newcommand{\Ccal}{\mathcal{C}}
\newcommand{\Dcal}{\mathcal{D}}
\newcommand{\Ecal}{\mathcal{E}}
\newcommand{\Fcal}{\mathcal{F}}
\newcommand{\Gcal}{\mathcal{G}}
\newcommand{\Hcal}{\mathcal{H}}
\newcommand{\Ical}{\mathcal{I}}
\newcommand{\Jcal}{\mathcal{J}}
\newcommand{\Kcal}{\mathcal{K}}
\newcommand{\Lcal}{\mathcal{L}}
\newcommand{\Mcal}{\mathcal{M}}
\newcommand{\Ncal}{\mathcal{N}}
\newcommand{\Ocal}{\mathcal{O}}
\newcommand{\Pcal}{\mathcal{P}}
\newcommand{\Qcal}{\mathcal{Q}}
\newcommand{\Rcal}{\mathcal{R}}
\newcommand{\Scal}{\mathcal{S}}
\newcommand{\Tcal}{\mathcal{T}}

\newcommand{\Ascr}{\mathscr{A}}
\newcommand{\Bscr}{\mathscr{B}}
\newcommand{\Cscr}{\mathscr{C}}
\newcommand{\Dscr}{\mathscr{D}}
\newcommand{\Escr}{\mathscr{E}}
\newcommand{\Fscr}{\mathscr{F}}
\newcommand{\Gscr}{\mathscr{G}}
\newcommand{\Hscr}{\mathscr{H}}
\newcommand{\Iscr}{\mathscr{I}}
\newcommand{\Jscr}{\mathscr{J}}
\newcommand{\Kscr}{\mathscr{K}}
\newcommand{\Lscr}{\mathscr{L}}
\newcommand{\Mscr}{\mathscr{M}}
\newcommand{\Nscr}{\mathscr{N}}
\newcommand{\Oscr}{\mathscr{O}}
\newcommand{\Pscr}{\mathscr{P}}
\newcommand{\Qscr}{\mathscr{Q}}
\newcommand{\Rscr}{\mathscr{R}}
\newcommand{\Sscr}{\mathscr{S}}
\newcommand{\Tscr}{\mathscr{T}}



\newcommand{\Ffrak}{\mathfrak{F}}
\newcommand{\Hfrak}{\mathfrak{H}}
\newcommand{\Lfrak}{\mathfrak{L}}

\newcommand{\zerott}{\mathtt{0}}
\newcommand{\onett}{\mathtt{1}}



%Lowercase Bold Letters
\newcommand{\bfa}{\mathbf{a}}
\newcommand{\bfb}{\mathbf{b}}
\newcommand{\bfc}{\mathbf{c}}
\newcommand{\bfd}{\mathbf{d}}
\newcommand{\bfe}{\mathbf{e}}
\newcommand{\bff}{\mathbf{f}}
\newcommand{\bfg}{\mathbf{g}}
\newcommand{\bfh}{\mathbf{h}}
\newcommand{\bfi}{\mathbf{i}}
\newcommand{\bfj}{\mathbf{j}}
\newcommand{\bfk}{\mathbf{k}}
\newcommand{\bfl}{\mathbf{l}}
\newcommand{\bfm}{\mathbf{m}}
\newcommand{\bfn}{\mathbf{n}}
\newcommand{\bfo}{\mathbf{o}}
\newcommand{\bfp}{\mathbf{p}}
\newcommand{\bfq}{\mathbf{q}}
\newcommand{\bfr}{\mathbf{r}}
\newcommand{\bfs}{\mathbf{s}}
\newcommand{\bft}{\mathbf{t}}
\newcommand{\bfu}{\mathbf{u}}
\newcommand{\bfv}{\mathbf{v}}
\newcommand{\bfw}{\mathbf{w}}
\newcommand{\bfx}{\mathbf{x}}
\newcommand{\bfy}{\mathbf{y}}
\newcommand{\bfz}{\mathbf{z}}


\newcommand{\id}{\text{id}}
\newcommand{\wt}{\text{wt}}
\newcommand{\puzzles}{\text{-puzzles}}



\newcommand{\inv}{^{-1}}





\def\IntKern{}
\newcommand\MySymbolint[2][0pt]{%
  \mathchoice
  {\def\IntKern{\dimexpr-.6\wd0+#1\relax}\Symbolint\displaystyle\displaystyle{\scalebox{.8}{$#2$}}}%
  {\def\IntKern{\dimexpr-.62\wd0+#1\relax}\Symbolint\textstyle\textstyle{\scalebox{.6}{$#2$}}}%
  {\def\IntKern{\dimexpr-.642\wd0+#1\relax}\Symbolint\scriptstyle\scriptscriptstyle{\scalebox{.45}{$#2$}}}%
 {\def\IntKern{\dimexpr-.67\wd0+#1\relax}\Symbolint\scriptscriptstyle\scriptscriptstyle{\scalebox{.35}{$#2$}}}%
  \!\int}
\newcommand\Symbolint[3]{%
  {\setbox0=\hbox{$#1{#2#3}{\int}$ }%
  \vcenter{\hbox{$#2#3$ }}\kern\IntKern}
}
%\def\uptriint{\MySymbolint\square}
\newcommand{\Kint}{\MySymbolint{\textnormal{K}}}



\DeclareMathOperator{\Span}{span}
\newcommand{\Gr}{\textnormal{Gr}}
\newcommand{\lam}{\lambda}
\newcommand{\GL}{\textnormal{GL}}
\newcommand{\Fl}{F\ell}
\DeclareMathOperator{\im}{Im}
\DeclareMathOperator{\codim}{codim}
\DeclareMathOperator{\rowspan}{rowspan}
\DeclareMathOperator{\sort}{sort}
\DeclareMathOperator{\content}{content}
\newcommand{\ano}{a_0+a_1}
\newcommand{\lno}{l_0+l_1}
\newcommand{\FF}{F_\bullet}
\newcommand{\GF}{G_\bullet}
\newcommand{\HF}{H_\bullet}
\newcommand{\VF}{V_F}
\newcommand{\VH}{V_H}
\newcommand{\pt}{\text{pt}}

\newcommand{\north}{\textnormal{N}}
\newcommand{\south}{\textnormal{S}}
\newcommand{\east}{\textnormal{E}}
\newcommand{\west}{\textnormal{W}}
\newcommand{\NE}{\textnormal{NE}}
\newcommand{\NW}{\textnormal{NW}}
\newcommand{\SE}{\textnormal{SE}}
\newcommand{\SW}{\textnormal{SW}}


%Font for fillings/labelings/conditions
\newcommand{\Ffil}{\mathtt{F}}
\newcommand{\Hfil}{\mathtt{H}}
\newcommand{\Kfil}{\mathtt{K}}
\newcommand{\Lfil}{\mathtt{L}}


%notation for strings
\newcommand{\zerostr}{\mathtt{0}}
\newcommand{\onestr}{\mathtt{1}}
\newcommand{\tenstr}{\mathtt{1\! 0}}
\newcommand{\twostr}{\mathtt{2}}
\newcommand{\threestr}{\mathtt{3}}
\newcommand{\dstr}{\mathtt{d}}
\newcommand{\mstr}{\mathtt{m}}

%notation for padding out a string
%\newcommand{\pad}[1]{\overline{#1}}
\newcommand{\pad}{^!}




\newcommand{\Hpieces}{H}   %notation for H* puzzle pieces

\newcommand{\flag}[1]{#1_\bullet}
\newcommand{\Ftilde}{\Tilde{F}}






%PUZZLE PIECES
\newcommand{\cccdeltapiece}{\raisebox{-.4\height}{\input{cccdeltapiece}}}

\newcommand{\cccnablapiece}{\raisebox{-.4\height}{\input{cccnablapiece}}}

\newcommand{\eqvtpiece}{\raisebox{-.4\height}{\input{eqvtpiece}}}



%BOUNDARY SHAPE SYMBOLS
\newcommand{\inlinetri}{
\tikzset{every picture/.style={line width=0.75pt}} %set default line width to 0.75pt        
\begin{tikzpicture}[x=0.75pt,y=0.75pt,yscale=-1,xscale=1,scale=.65]
%uncomment if require: \path (0,300); %set diagram left start at 0, and has height of 300
%Shape: Polygon [id:ds864980626463701] 
\raisebox{-.1ex}
{
\draw  [line width=0.75]  (18.6,6.68) -- (28.6,24) -- (8.6,24) -- cycle ;
}
\end{tikzpicture}
}


\newcommand{\inlinetrap}{
\tikzset{every picture/.style={line width=0.75pt}} %set default line width to 0.75pt  
\begin{tikzpicture}[x=0.75pt,y=0.75pt,yscale=-1,xscale=1,scale=.75]
%uncomment if require: \path (0,300); %set diagram left start at 0, and has height of 300
%Shape: Polygon [id:ds4037661691821606] 
\raisebox{-.2ex}
{
\draw   (34.67,22.4) -- (29.67,13.74) -- (19.67,13.74) -- (14.67,22.4) -- cycle ;
}
\end{tikzpicture}
}

\newcommand{\inlinetrapside}{
\tikzset{every picture/.style={line width=0.75pt}} %set default line width to 0.75pt  
\begin{tikzpicture}[x=0.75pt,y=0.75pt,yscale=-1,xscale=1,scale=.75]
%uncomment if require: \path (0,300); %set diagram left start at 0, and has height of 300
%Shape: Polygon [id:ds1877413286993055] 
\raisebox{-.35ex}
{
\draw   (22.8,24) -- (12.8,24) -- (7.8,15.34) -- (12.8,6.68) -- cycle ;
}
\end{tikzpicture}
}

\newcommand{\inlinetrapsiderotate}{
\tikzset{every picture/.style={line width=0.75pt}} %set default line width to 0.75pt        
\begin{tikzpicture}[x=0.75pt,y=0.75pt,yscale=-1,xscale=1,scale=.75]
%uncomment if require: \path (0,300); %set diagram left start at 0, and has height of 300
%Shape: Boxed Polygon [id:dp23745649106798106] 
\raisebox{-.5ex}
{
\draw   (11.8,10.68) -- (21.8,10.68) -- (26.8,19.34) -- (21.8,28) -- cycle ;
}
\end{tikzpicture}}

\newcommand{\inlinetrapsiderotatetwo}{
\tikzset{every picture/.style={line width=0.75pt}} %set default line width to 0.75pt  
\begin{tikzpicture}[x=0.75pt,y=0.75pt,yscale=-1,xscale=1,scale=.75]
%uncomment if require: \path (0,300); %set diagram left start at 0, and has height of 300
%Shape: Polygon [id:ds1877413286993055] 
\raisebox{-.35ex}
{
\draw   (28.05,9.35) -- (33.05,18.01) -- (28.05,26.67) -- (18.05,26.67) -- cycle ;
}
\end{tikzpicture}
}

\newcommand{\inlinepar}{
\tikzset{every picture/.style={line width=0.75pt}} %set default line width to 0.75pt        
\begin{tikzpicture}[x=0.75pt,y=0.75pt,yscale=-1,xscale=1,scale=.65]
%uncomment if require: \path (0,300); %set diagram left start at 0, and has height of 300
%Shape: Polygon [id:ds43109096202440855] 
\raisebox{-.7ex}
{
\draw  [line width=0.75]  (14.4,11.02) -- (24.4,28.34) -- (19.4,37) -- (9.4,19.68) -- cycle ;
}
\end{tikzpicture}
}

\newcommand{\inlinerhombus}{
\tikzset{every picture/.style={line width=0.75pt}} %set default line width to 0.75pt        
\begin{tikzpicture}[x=0.75pt,y=0.75pt,yscale=-1,xscale=1,scale=.5]
%uncomment if require: \path (0,61); %set diagram left start at 0, and has height of 61
%Shape: Polygon [id:ds13444390995271494] 
\raisebox{-.7ex}
{
\draw   (23.33,8.96) -- (33.33,26.28) -- (23.33,43.6) -- (13.33,26.28) -- cycle ;
}
\end{tikzpicture}
}

\newcommand{\inlinepent}{
\tikzset{every picture/.style={line width=0.75pt}} %set default line width to 0.75pt        
\begin{tikzpicture}[x=0.75pt,y=0.75pt,yscale=-1,xscale=1,scale=.55]
%uncomment if require: \path (0,300); %set diagram left start at 0, and has height of 300
%Shape: Polygon [id:ds5570960031147962] 
\raisebox{-.5ex}
{
\draw   (22,8.02) -- (32,25.34) -- (27,34) -- (17,34) -- (12,25.34) -- cycle ;
}
\end{tikzpicture}}

\newcommand{\inlinepentside}{
\tikzset{every picture/.style={line width=0.75pt}} %set default line width to 0.75pt        
\begin{tikzpicture}[x=0.75pt,y=0.75pt,yscale=-1,xscale=1,scale=.6]
%uncomment if require: \path (0,300); %set diagram left start at 0, and has height of 300
%Shape: Boxed Polygon [id:dp9392694177909653] 
\raisebox{-.3ex}
{
\draw   (32.75,35.5) -- (42.75,18.18) -- (52.75,18.18) -- (57.75,26.84) -- (52.75,35.5) -- cycle ;
}
\end{tikzpicture}}

\newcommand{\inlinehex}{
\tikzset{every picture/.style={line width=0.75pt}} %set default line width to 0.75pt        
\begin{tikzpicture}[x=0.75pt,y=0.75pt,yscale=-1,xscale=1,scale=.65]
%uncomment if require: \path (0,300); %set diagram left start at 0, and has height of 300
%Shape: Polygon [id:ds18294503666304496] 
\raisebox{-.3ex}
{
\draw  [line width=0.75]  (24,9.68) -- (29,18.34) -- (24,27) -- (14,27) -- (9,18.34) -- (14,9.68) -- cycle ;
}
\end{tikzpicture}
}


%where the tikz stuff ends





%% define your title in the usual way
\title[]{Commutative Properties of Schubert Puzzles with Convex Polygonal Boundary Shapes}

%% define your authors in the usual way
%% use \addressmark{1}, \addressmark{2} etc for the institutions, and use \thanks{} for contact details
%% for more than 3 names, use the Oxford comma: One Author, Two Author, Red Author, \and Blue Author
\author[]{ \texorpdfstring{Portia X. Anderson\addressmark{1} }{Portia X. Anderson } } %\thanks{\href{mailto:hello@world.c}{hello@world.c}. Longer names me was partially supported by Grant 2017.11.14.$\partial$\;supp.}
%\and you\addressmark{2}


%% then use \addressmark to match authors to institutions here
\address{\addressmark{1}Department of Mathematics, Cornell University, Ithaca, NY 
%\\ \addressmark{2}School of Something, University of Something Else, Somewhere
}

%% put the date of submission here
\received{November 15, 2024}

%% leave this blank until submitting a revised version
\revised{April 1, 2025}

%% put your English abstract here, or comment this out if you don't have one yet
%% please don't use custom commands in your abstract / resume, as these will be displayed online
%% likewise for citations -- please don't use \cite, and instead write out your citation as something like (author year)
\abstract{We generalize classical triangular Schubert puzzles to puzzles with convex polygonal boundary. We give these puzzles a geometric Schubert calculus interpretation and derive novel combinatorial commutativity statements, using both geometric and combinatorial arguments, for puzzles with four, five, and six sides, having various types of symmetry in their boundary conditions.}

%% put your French abstract here, or comment this out if you don't have one
%\resume{\lipsum[2]}

%% put your keywords here, or comment this out if you don't have them yet
\keywords{Schubert calculus, Schubert puzzle}

%% you can include your bibliography however you want, but using an external .bib file is STRONGLY RECOMMENDED and will make the editor's life much easier
%% regardless of how you do it, please use numerical citations; i.e., [xx, yy] in the text

%% this sample uses biblatex, which (among other things) takes care of URLs in a more flexible way than bibtex
%% but you can use bibtex if you want
\usepackage[backend=bibtex, giveninits]{biblatex}
\addbibresource{bibliography.bib}
%% note the \printbibliography command at the end of the file which goes with these biblatex commands

\begin{document}

\maketitle
%% note that you DO NOT have to put your abstract here -- it is generated by \maketitle and the \abstract and \resume commands above

\section{Introduction}


Since their introduction in \cite{KTW} in 1999, Schubert puzzles have been the subject of continuous research activity, but little work has been done to study non-triangular puzzles up until now. In this extended abstract, we give a survey of the commutative properties of generalized convex polygonal Schubert puzzles of the type that compute structure constants in the ordinary cohomology of the Grassmannian. These properties are analogous to the commutative property of classical triangular puzzles. We remark that these results extend to puzzles that compute the cohomology of 2-step and 3-step flag varieties, as well as puzzles that compute the K-theory of the Grassmannian. We also present an analogue of commutativity for parallelogram-shaped equivariant puzzles.

This work was motivated by the observation that convex polygonal puzzles can be given a geometric interpretation by associating them to classes of triangular puzzles whose boundary conditions follow a certain form. Once translated to this more classical setting, the statements about polygonal puzzles can be proven geometrically or by using basic properties of triangular puzzles. Future research directions include finding interpretations or extensions of these results in other contexts.






\subsection{Schubert Puzzles}

\subsubsection{Classical triangular puzzles}


Classically, a \textbf{Schubert puzzle} is a tiling of an equilateral triangular region, whose boundary we denote $\inlinetri$, using a set  of labeled unit triangles 
\begin{center}
\input{puzzlepieceset}
\end{center}
called \textbf{puzzle pieces}, %so that the edge labels match up along all glued edges, 
so that any two glued edges have the same label, and only $\zerostr$ and $\onestr$ labels appear along the outer boundary $\inlinetri$, not $\tenstr$s. %We denote this puzzle piece set as $\Hfrak$.


An equilateral triangular boundary whose NW, NE, and South sides are labeled with strings $\lambda$, $\mu$, and $\nu$ in the orientations shown in \Cref{fig:lambdamunuboundary} will be denoted $\inlinetri_{\lambda,\mu,\nu}$, and a puzzle with this boundary labeling will be called a \textbf{$\inlinetri_{\lambda,\mu,\nu}$-puzzle}. See \Cref{fig:1010_0101_0011} for an example.

\begin{figure}[ht]
\centering
    \begin{subfigure}[t]{0.4\textwidth}
        \centering
        \input{lambdamunuboundary}
        \caption{The labeled boundary $\protect\inlinetri_{\lambda,\mu,\nu}$}
        \label{fig:lambdamunuboundary}
    \end{subfigure}
    ~
    \begin{subfigure}[t]{0.4\textwidth}
        \centering
        \input{1010_0101_0011}
        
        \caption{A $\protect\inlinetri_{\onestr\zerostr\onestr\zerostr,\zerostr\onestr\zerostr\onestr,\zerostr\zerostr\onestr\onestr}$-puzzle}
        \label{fig:1010_0101_0011}
    \end{subfigure}
    \caption{}
    \label{fig:triangularboundaryandpuzzle}
\end{figure}


Triangular puzzles have three-fold rotational symmetry, i.e. there are bijections 
$$\{\inlinetri_{\lambda,\mu,\nu}\text{-puzzles}\}\leftrightarrow \{\inlinetri_{\mu,\nu,\lambda}\text{-puzzles}\} \leftrightarrow \{\inlinetri_{\nu,\lambda,\mu}\text{-puzzles}\}$$ obtained by rotating puzzles in $120^\circ$ increments. Triangular puzzles also have a commutative property, i.e. that 
$\#\{\inlinetri_{\lambda,\mu,\nu}\text{-puzzles}\}= \#\{\inlinetri_{\mu,\lambda,\nu}\text{-puzzles}\}$, which is non-obvious from a combinatorial standpoint. A bijective combinatorial proof is given by Purbhoo in \cite{Purbhoo} using mosaics, which are objects that are in natural bijection with puzzles.



\subsubsection{Equivariant puzzles}
An \textbf{equivariant puzzle}, first described in \cite{KTpuzzles}, is one in which we additionally allow the \textit{equivariant piece} \eqvtpiece. This piece has a \textit{weight} of the form $y_j-y_i$, where $(i,j)$ is uniquely determined by its position in the puzzle.

\subsubsection{Puzzles with convex polygonal boundary}

Now we will generalize the definition of ``puzzles'' to include puzzle piece tilings of convex polygonal shapes, where again only $\zerostr$ and $\onestr$ labels are allowed to appear along the outer boundary, not $\tenstr$s. 
%\begin{enumerate}
    %\item the angle of each puzzle piece edge relative to the $x$-axis is a multiple of $60^\circ$, and 
    %\item only $\zerostr$ and $\onestr$ labels are allowed to appear along the outer boundary, not $\tenstr$s.
%\end{enumerate}
This allows us to have puzzles with trapezoidal, parallelogram-shaped, pentagonal, and hexagonal boundary as well. (See \Cref{fig:boundaryshapespuzzles} for examples.)
%In fact, a hexagon generalizes all these shapes; a boundary with three, four, or five sides is just a hexagon where some sides have length 0.

Similarly to $\inlinetri_{\lambda,\mu,\nu}$, we will use a shape symbol (where only the number of sides and angles matter, not side lengths) with a subscript sequence of strings, read clockwise starting from the SW, to denote a labeled boundary.
%, for example  $\inlinetrapside_{\lambda,\mu,\nu,\xi}$, $\inlinepar_{\lambda,\mu,\nu,\xi}$, $\inlinepent_{\lambda,\mu,\nu,\xi,o}$, or $\inlinehex_{\lambda,\mu,\nu,\xi,o,\pi}$. 
%For every shape, we will always read the sequence of labels beginning at the SW side and traveling clockwise around the boundary. 
%always read the labels traveling clockwise around the boundary, 
%For any shape, we will always begin at the SW side of the boundary and read the sequence of labels, assigning one to each side of the boundary, in a clockwise orientation. 
%We will also use the suffix \textbf{-puzzle} to refer to a puzzle with the indicated boundary (see \Cref{fig:boundaryshapespuzzles}).

\begin{figure}[ht]
\centering
\begin{subfigure}[t]{0.25\textwidth}
\centering
\input{trapezoidexample}
\caption{$ \protect\inlinetrapside_{\onestr\zerostr\onestr,\zerostr\onestr\zerostr\onestr,\zerostr\zerostr\onestr\zerostr\onestr\onestr\onestr,\onestr\zerostr\onestr\zerostr}$-puzzle example}
\label{fig:trapezoidexample}
\end{subfigure}
\begin{subfigure}[t]{0.215\textwidth}
\centering
\input{parallelogramexample}
\caption{$ \protect\inlinepar_{\onestr\zerostr\onestr,\zerostr\onestr\zerostr\onestr,\zerostr\onestr\onestr,\zerostr\zerostr\onestr\onestr}$-puzzle example}
\label{fig:parallelogramexample}
\end{subfigure}
\begin{subfigure}[t]{0.235\textwidth}
\centering
\input{pentagonexample}
\caption{$ \protect\inlinepent_{\zerostr\zerostr,\zerostr\onestr\onestr\onestr\onestr,\zerostr\zerostr\zerostr\onestr,\onestr\onestr\onestr,\onestr\zerostr}$-puzzle example}
\label{fig:pentagonexample}
\end{subfigure}
\begin{subfigure}[t]{0.24\textwidth}
\centering
\input{hexagonexample}
\caption{$ \protect\inlinehex_{\zerostr,\zerostr\onestr\onestr\onestr,\zerostr\onestr,\zerostr\onestr,\zerostr\onestr\onestr,\zerostr\onestr\onestr}$-puzzle example}
\label{fig:hexagonexample}
\end{subfigure}

\centering
\caption{Examples of puzzles with convex polygonal boundary shapes.
%Examples of puzzles with different boundary shapes. 
%The boundary labels are read clockwise starting from the SW side.
}
\label{fig:boundaryshapespuzzles}
\end{figure}






\subsection{Schubert calculus}



\subsubsection{Schubert varieties}
\label{section:schubertvarieties}
Let $\Gr(k;\C^n)$ denote the \textbf{Grassmannian} of $k$-dimensional subspaces of $\C^n$. 
%It is a smooth complex manifold and projective algebraic variety of dimension $k(n-k)$.
We will write $\lambda\in\binom{[n]}{k}$ to mean that $\lambda$ is a \textbf{binary string} with a \textbf{content} of $k$ $\onestr$s and $n-k$ $\zerostr$s. Let $\lambda=\lambda_1\lambda_2\cdots\lambda_n\in\binom{[n]}{k}$, and let $F_\bullet$ be the standard complete flag $\{\mathbf{0}\}=F_0\subset F_1\subset\cdots\subset F_n=\C^n$. Then $X_\lambda$ is the \textbf{Schubert variety} in $\Gr(k;\C^n)$ defined by 
$$X_\lambda(F_\bullet):= \{V\in\Gr(k;\C^n) : \dim(V\cap F_i)\geq \lambda_1+\lambda_2+\cdots+\lambda_i\}.$$
We will use the superscript $^\vee$ to denote the operation of reversing a string. Then $X^\lambda$ is the \textbf{opposite Schubert variety} defined by $X^\lambda:=w_0\cdot X_{\lambda^\vee}$, where $w_0$ is the anti-diagonal permutation matrix.
The classes $\{[X_\lambda] : \lambda \in \binom{[n]}{k}\}$ form a $\Z$-basis for the cohomology ring $H^*(\Gr(k;\C^n))$, and
%So we can expand a product $[X_\lambda][X_\mu]$ uniquely as a linear combination
%$$[X_\lambda][X_\mu]=\sum_{\nu\in \binom{[n]}{k}}c_{\lambda,\mu}^\nu [X_\nu].$$
 the classes $\{[X^\lambda] : \lambda\in\binom{[n]}{k}\}$ form a dual basis with
$\int_{\Gr(k;\C^n)} [X_\lambda][X^\mu] = \delta_{\lambda,\mu}$.
%hence we have $c_{\lambda,\mu}^\nu= \underset{\Gr(k;\C^n)}{\int} [X_\lambda][X_\mu][X^\nu].$
The classes of these Schubert varieties also form dual bases over $\Z[y_1,\ldots,y_n]$ in the $T$-equivariant cohomology $H^*_T(\Gr(k;\C^n)):=H^*(ET\times^T\Gr(k;\C^n))$, where $T=(\C^\times)^n$ and $ET$ is a contractible space with a free $T$-action.

 \subsubsection{Schubert calculus}
 
The goal of \textbf{Schubert calculus} is to compute the structure constants in various cohomology theories in the Schubert basis. In $H^*(\Gr(k;\C^n))$, the  structure constants $c_{\lambda,\mu}^\nu= \int_{\Gr(k;\C^n)} [X_\lambda][X_\mu][X^\nu]$ are positive, and in fact they are the \textit{Littlewood--Richardson (LR) numbers} that arise in symmetric function theory.  (Usually these are indexed by triples of partitions, but we can identify our binary strings with partitions that fit in a $k\times (n-k)$ rectangle via a simple bijection.)  %We may also compute the structure constants in other cohomology theories, such as $H^*_T(\Gr(k;\C^n))$.



\subsubsection{Puzzles compute Schubert calculus}

It is a result of Knutson--Tao--Woodward in \cite{KTW} that puzzles using the basic puzzle piece set compute the structure constants of $H^*(\Gr(k;\C^n))$ in the Schubert basis:
%For $\lambda,\mu,\nu\in \binom{[n]}{k}$, we have
$$c_{\lambda,\mu}^\nu  = \#\{\inlinetri_{\lambda,\mu,\nu^\vee}\text{-puzzles}\}.$$

Knutson--Tao later proved in \cite{KTpuzzles} that equivariant puzzles compute the structure constants, which we denote $(c_T)_{\lambda,\mu}^\nu$, of $H^*_T(\Gr(k;\C^n))$ according to the following formula:
$$(c_T)_{\lambda,\mu}^\nu = \sum_{\inlinetri_{\lambda,\mu,\nu^\vee}\text{-puzzles } P} \text{weight}(P) = \sum_{\inlinetri_{\lambda,\mu,\nu^\vee}\text{-puzzles } P} \left( \prod_{\substack{\text{equivariant} \\ \text{pieces }p \text{ in } P }} \text{weight}(p) \right).$$ 




\subsection{Main Theorems}


%In the following set of theorems, we generalize the commutative property of classical triangular puzzles to puzzles with polygonal boundary shapes. 


\begin{theorems}[Commutative Properties of Convex Polygonal Puzzles] 
\label{thms:commutativeproperties} For each labeled boundary shape drawn below, we can commute the labels on any pair of sides with matching colored squiggly lines while preserving the number of puzzles filling the boundary.
Namely,

\begin{enumerate}[label=(\alph*)]
    \item 
    \begin{minipage}{.75\linewidth}
    \centering
        (Commutative Property of Puzzles with Split Symmetry)
    
         For $\alpha,\beta\in\binom{[a]}{a_1}$, $\gamma,\delta\in\binom{[c]}{c_1}$, and $\nu\in\binom{[c+a]}{a_1+c_1}$, 
    \begin{gather*}
        \#\{\inlinetri_{\alpha\gamma,\nu,\delta\beta} \text{-puzzles}\}= \#\{\inlinetri_{\beta\gamma,\nu,\delta\alpha} \text{-puzzles}\}\\
        = \#\{\inlinetri_{\alpha\delta,\nu,\gamma\beta} \text{-puzzles}\} = \#\{\inlinetri_{\beta\delta,\nu,\gamma\alpha} \text{-puzzles}\}.
        \end{gather*}
    \end{minipage}
   \begin{minipage}{.23\linewidth}
   \centering
        \input{triangleboundarycommutesplit}
    \end{minipage}
    
    \hrulefill
    
    \item 
    \begin{minipage}{.75\linewidth}
    \centering
        (Commutative Property of Trapezoidal Puzzles)
    
         For $\beta\in\binom{[a]}{a_1}$, $\gamma,\delta\in\binom{[c]}{c_1}$, and $\nu\in\binom{[c+a]}{a_1+c_1}$, 
    \begin{gather*}
        \#\{\inlinetrapside_{\beta,\gamma,\nu,\delta} \text{-puzzles}\} = \#\{\inlinetrapside_{\beta,\delta,\nu,\gamma} \text{-puzzles}\}.
        \end{gather*}
    \end{minipage}
   \begin{minipage}{.23\linewidth}
   \centering
         \input{trapezoidboundarycommute} 
    \end{minipage}

    \hrulefill

    \item 
    \begin{minipage}{.75\linewidth}
    \centering
          (Commutative Property of Parallelogram-Shaped Puzzles)
          
          For $\alpha,\beta\in\binom{[a]}{a_1}$ and $\gamma,\delta\in\binom{[c]}{c_1}$, 
    \begin{gather*}
        \#\{\inlinepar_{\alpha,\gamma,\beta,\delta} \text{-puzzles}\} =  \#\{\inlinepar_{\beta,\gamma,\alpha,\delta} \text{-puzzles}\} \\
        =  \#\{\inlinepar_{\alpha,\delta,\beta,\gamma} \text{-puzzles}\} =  \#\{\inlinepar_{\beta,\delta,\alpha,\gamma} \text{-puzzles}\}.
        \end{gather*}
    \end{minipage}
   \begin{minipage}{.23\linewidth}
   \centering
         \input{parallelogramboundarycommute}
    \end{minipage}

    \hrulefill

    \item 
    \begin{minipage}{.75\linewidth}
    \centering
          (Commutative Property of Symmetric Rhombus-Shaped Puzzles)
          
          For $\alpha,\beta,\gamma,\delta\in\binom{[a]}{a_1}$ and any bijection $f:\{\alpha,\beta,\gamma,\delta\} \rightarrow \{\alpha,\beta,\gamma,\delta\}$, 
    \begin{gather*}
        \#\{\inlinerhombus_{\alpha,\gamma,\beta,\delta} \text{-puzzles}\} =  \#\{\inlinerhombus_{f(\alpha),f(\gamma),f(\beta),f(\delta)} \text{-puzzles}\}.
        \end{gather*}
    \end{minipage}
   \begin{minipage}{.23\linewidth}
   \centering
         \input{rhombusboundarycommute}
    \end{minipage}

    \hrulefill

    \item 
    \begin{minipage}{.75\linewidth}
    \centering
            (Commutative Property of Symmetric Pentagonal Puzzles)
    
          For $\beta,\zeta\in\binom{[b]}{b_1}$, $\gamma,\epsilon\in\binom{[c]}{c_1}$, and $\delta\in\binom{[d]}{d_1}$, 
    \begin{gather*}
        \#\{\inlinepentside_{\beta,\gamma,\delta,\epsilon,\zeta} \text{-puzzles}\} = \#\{\inlinepentside_{\zeta,\gamma,\delta,\epsilon,\beta} \text{-puzzles}\} \\
        = \#\{\inlinepentside_{\beta,\epsilon,\delta,\gamma,\zeta} \text{-puzzles}\} = \#\{\inlinepentside_{\zeta,\epsilon,\delta,\gamma,\beta} \text{-puzzles}\}.
        \end{gather*}
    \end{minipage}
   \begin{minipage}{.23\linewidth}
   \centering
         \input{pentagonboundarycommute}
    \end{minipage}

    \hrulefill

    \item 
    \begin{minipage}{.75\linewidth}
    \centering
            (Commutative Property of Hexagonal Puzzles with Opposite Sides Symmetry)
        
          For $\alpha,\delta\in\binom{[a]}{a_1}$, $\beta,\epsilon\in\binom{[b]}{b_1}$, and $\gamma,\zeta\in\binom{[c]}{c_1}$, and any bijections $f:\{\alpha,\delta\} \rightarrow \{\alpha,\delta\}$, $g:\{\beta,\epsilon\} \rightarrow \{\beta,\epsilon\}$, and $h:\{\gamma,\zeta\} \rightarrow \{\gamma,\zeta\}$, 
    \begin{gather*}
        \#\{\inlinehex_{\alpha,\beta,\gamma,\delta,\epsilon,\zeta} \text{-puzzles}\} = \#\{\inlinehex_{f(\alpha),g(\beta),h(\gamma),f(\delta),g(\epsilon),h(\zeta)} \text{-puzzles}\}.
        \end{gather*}
    \end{minipage}
   \begin{minipage}{.23\linewidth}
   \centering
         \input{hexagonboundarycommuteopposite}
    \end{minipage}

    \hrulefill

    \item 
    \begin{minipage}{.75\linewidth}
    \centering
        (Commutative Property of Hexagonal Puzzles with Two-Way Symmetry)
    
          For $\alpha\in\binom{[a]}{a_1}$, $\beta,\zeta\in\binom{[b]}{b_1}$, $\gamma,\epsilon\in\binom{[c]}{c_1}$, and $\delta\in\binom{[d]}{d_1}$, 
    \begin{gather*}
        \#\{\inlinehex_{\alpha,\beta,\gamma,\delta,\epsilon,\zeta} \text{-puzzles}\} = \#\{\inlinehex_{\alpha,\zeta,\gamma,\delta,\epsilon,\beta} \text{-puzzles}\}\\
        = \#\{\inlinehex_{\alpha,\beta,\epsilon,\delta,\gamma,\zeta} \text{-puzzles}\} = \#\{\inlinehex_{\alpha,\zeta,\epsilon,\delta,\gamma,\beta} \text{-puzzles}\}.
        \end{gather*}
    \end{minipage}
   \begin{minipage}{.23\linewidth}
   \centering
         \input{hexagonboundarycommutetwowayside}
    \end{minipage}

    \hrulefill

    \item 
    \begin{minipage}{.75\linewidth}
    \centering
            (Commutative Property of Hexagonal Puzzles with Three-Way Symmetry)
    
          For $\alpha,\gamma,\epsilon\in\binom{[a]}{a_1}$ and $\beta,\delta,\zeta\in\binom{[b]}{b_1}$, and any bijections $f:\{\alpha,\gamma,\epsilon\}\rightarrow \{\alpha,\gamma,\epsilon\}$ and $g:\{\beta,\delta,\zeta\}\rightarrow \{\beta,\delta,\zeta\}$, 
    \begin{gather*}
        \#\{\inlinehex_{\alpha,\beta,\gamma,\delta,\epsilon,\zeta} \text{-puzzles}\} = \#\{\inlinehex_{f(\alpha),g(\beta),f(\gamma),g(\delta),f(\epsilon),g(\zeta)} \text{-puzzles}\}.
        \end{gather*}
    \end{minipage}
   \begin{minipage}{.23\linewidth}
   \centering
         \input{hexagonboundarycommutethreeway}
    \end{minipage}


    \hrulefill


    
    \item 
    \begin{minipage}{.75\linewidth}
    \centering
            (Commutative Property of Hexagonal Puzzles with All-Way Symmetry)
    
          For $\alpha,\beta,\gamma,\delta,\epsilon,\zeta\in\binom{[a]}{a_1}$ and any bijection $f:\{\alpha,\beta,\gamma,\delta,\epsilon,\zeta\}\rightarrow \{\alpha,\beta,\gamma,\delta,\epsilon,\zeta\}$, 
    \begin{gather*}
        \#\{\inlinehex_{\alpha,\beta,\gamma,\delta,\epsilon,\zeta} \text{-puzzles}\} = \#\{\inlinehex_{f(\alpha),f(\beta),f(\gamma),f(\delta),f(\epsilon),f(\zeta)} \text{-puzzles}\}.
        \end{gather*}
    \end{minipage}
   \begin{minipage}{.23\linewidth}
   \centering
         \input{hexagonboundarycommuteallway}
    \end{minipage}
    
\end{enumerate}

\end{theorems}

\begin{remark}
    \normalfont \Cref{thms:commutativeproperties} have analogous statements for 2-step and 3-step puzzles, as well as puzzles that include the piece \cccdeltapiece, which compute stucture constants in the K-theory $K(\Gr(k;\C^n))$ in the Schubert structure sheaf basis $\{[\Ocal_\lambda]: \lambda\in\binom{[n]}{k}\}$. Most proofs go through with little change. Details are given in \cite{anderson2024} but are omitted here.
\end{remark}

In \Cref{section:commutativepropertyofeqvtparallelogram}, we will also give a commutative property for parallelogram-shaped equivariant puzzles.


\section{Preliminaries}



\subsection{A geometric interpretation of convex polygonal puzzles}

 We can give a convex polygonal puzzle a familiar geometric Schubert calculus interpretation via an operation to ``complete'' it to a triangular puzzle.

 Let \textbf{sort} denote the operation of moving all of the $\zerostr$s in a binary string ahead of all the $\onestr$s. So, for $\lambda\in\binom{[n]}{k}$, we have $\sort(\lambda)=\zerostr^{n-k}\onestr^{k}$.

 \begin{proposition}
     %For $\alpha,\beta\in\binom{[a]}{a_1}$, $\gamma,\delta\in\binom{[c]}{c_1}$, and $\nu\in\binom{[c+a]}{a_1+c_1}$, we have a bijection
    %The operations of ``completing to a triangle'' yield the following statements.
    By completing polygonal puzzles to triangular puzzles, we obtain the following:
     

     \begin{enumerate}[label=(\alph*)]
     
         \item For $\beta\in\binom{[a]}{a_1}$, $\gamma,\delta\in\binom{[c]}{c_1}$, and $\nu\in\binom{[c+a]}{a_1+c_1}$, there is a bijection 
         $$\{\inlinetrapside_{\beta,\gamma,\nu,\delta}\text{-puzzles}\} \leftrightarrow \{\inlinetri_{\sort(\beta)\gamma,\nu,\delta\beta} \text{-puzzles}\}.$$
         \item For $\alpha,\beta\in\binom{[a]}{a_1}$ and $\gamma,\delta\in\binom{[c]}{c_1}$, there is a bijection
         $$\{\inlinepar_{\alpha,\gamma,\beta,\delta} \text{-puzzles}\} \leftrightarrow \{\inlinetri_{\sort(\alpha)\gamma,\beta\sort(\delta),\delta\alpha} \text{-puzzles}\}.$$
         \item For $\alpha\in\binom{[a]}{a_1}$, $\beta\in\binom{[b]}{b_1}$, $\gamma\in\binom{[c]}{c_1}$, $\delta\in\binom{[d]}{d_1}$, $\epsilon\in\binom{[e]}{e_1}$, and $\zeta\in\binom{[z]}{z_1}$, there is a bijection 
         $$\{\inlinehex_{\alpha,\beta,\gamma,\delta,\epsilon,\zeta} \text{-puzzles}\} \leftrightarrow \{\inlinetri_{\sort(\alpha)\beta\gamma,\sort(\gamma)\delta\sort(\epsilon),\epsilon\zeta\alpha} \text{-puzzles}\}.$$
     \end{enumerate}
    
 \end{proposition}

 \begin{proof}
     See \Cref{fig:completetotriangle}. Given the labels on the triangular boundaries, there exists a unique filling of each grey triangular region with puzzle pieces, which forces the labels around the inner pink regions to replicate those for the polygons on the top row. 
In each case, a bijection is given by gluing on the uniquely filled grey regions, and the inverse is cutting them off.
 \end{proof}

\begin{figure}[ht]
\centering
\begin{subfigure}[t]{.31\textwidth}
\centering
    \scalebox{.75}{\input{trapezoidfilling}}
    
\Large \quad\, $\updownarrow$ \normalsize

\vspace{.2\baselineskip}

\scalebox{.75}{\input{trapezoidcomplete}}
\caption{
%Completing a trapezoidal puzzle to a triangular puzzle.
}
\label{fig:trapezoidcomplete}
\end{subfigure}
\begin{subfigure}[t]{.31\textwidth}
\centering
    \scalebox{.75}{\input{parallelogramfilling}}
    
\Large \, $\updownarrow$ \normalsize

\vspace{.2\baselineskip}

\scalebox{.75}{\input{parallelogramcomplete}}
    \caption{
    %Completing a parallelogram-shaped puzzle to a triangular puzzle.
    }
\label{fig:parallelogramcomplete}
\end{subfigure}
\begin{subfigure}[t]{.34\textwidth}
\centering
    \scalebox{.75}{\input{hexagonfilling}}
    
\!\! \Large $\updownarrow$ \normalsize

\scalebox{.75}{\input{hexagoncomplete}}
    \caption{
    %Completing a hexagonal puzzle to a triangular puzzle.
    }
\label{fig:hexagoncomplete}
\end{subfigure}
\caption{
The operations of completing polygonal puzzles to triangular puzzles.
}
\label{fig:completetotriangle}
\end{figure}




\subsection{Geometric lemmas}

The following two lemmas form the foundation upon which all of the geometric proofs of the theorems are built. In particular, \Cref{lemma:commutesplit} tells us how we can swap substrings in the labels on two different sides of a triangular boundary when certain symmetries are present, such as the ones seen in the triangular boundary labels in \Cref{fig:completetotriangle}.


\begin{lemma}
\label{lemma:directsum}
Let $\flag{F}$ and $\flag{\Ftilde}$ be the standard and anti-standard complete flags in $\C^{a+c}$, respectively. Let $T:=(\C^\times)^{a+c}$, and define the $T$-equivariant closed immersion
$$\Omega: \Gr(a_1;F_a)\times \Gr(c_1;\Ftilde_c) \hookrightarrow  \Gr(a_1+c_1;\C^{a+c}), \quad (V,W)\mapsto V\oplus W.$$ 
Then for $\alpha,\beta\in\binom{[a]}{a_1}$ and $\gamma,\delta\in\binom{[c]}{c_1}$, we have
\begin{equation*}
    X_{\alpha\gamma}\cap X^{(\delta\beta)^\vee} = \Omega((X_\alpha\cap X^{\beta^\vee}) \times (X_\gamma\cap X^{\delta^\vee})), \tag{1}
\end{equation*}
and 
\begin{equation*}
[X_{\alpha\gamma}][X^{(\delta\beta)^\vee}]= \Omega_*([X_\alpha][X^{\beta^\vee}]\otimes [X_\gamma][X^{\delta^\vee}]) \tag{2}
\end{equation*}
in $H_T^*(\Gr(a_1+c_1;\C^{a+c}))$ and in $H^*(\Gr(a_1+c_1;\C^{a+c}))$.



\end{lemma}



\begin{proof}[Proof idea] In \cite{anderson2024}, we  prove the equality of sets in statement (1) by checking that there is an inclusion in both directions, by directly verifying that the conditions on intersection dimensions encoded by the strings are satisfied. Then statement (2) follows immediately from statement (1).
\end{proof}



\begin{lemma}
\label{lemma:commutesplit}

Define block matrices
$\Phi_a:=\left[
\begin{array}{c|c}
J_{a} & \mathbf{0} \\ \hline
\mathbf{0} & I_{c}
\end{array}\right]$
and
$\Phi_c:=\left[
\begin{array}{c|c}
I_{a} & \mathbf{0} \\ \hline
\mathbf{0} & J_{c}
\end{array}\right]$ 
in $\GL(\C^{a+c})$, where $I_a$ and $J_a$ (resp. $I_c$ and $J_c$) denote the $a\times a$ (resp. $c\times c$) identity and anti-diagonal permutation matrices, respectively.
Let $\alpha,\beta\in\binom{[a]}{a_1}$ and $\gamma,\delta\in\binom{[c]}{c_1}$. Then, 

\begin{enumerate}[label=(\roman*)]
    \item in $H_T^*(\Gr(a_1+c_1;\C^{a+c}))$, we have 
$$[X_{\alpha\gamma}][X^{(\delta\beta)^\vee}]= \Phi_a\cdot ([X_{\beta\gamma}][X^{(\delta\alpha)^\vee}]) = \Phi_c\cdot ([X_{\alpha\delta}][X^{(\gamma\beta)^\vee}]) = \Phi_c\cdot\Phi_a\cdot ([X_{\beta\delta}][X^{(\gamma\alpha)^\vee}])$$
    \item in $H^*(\Gr(a_1+c_1;\C^{a+c}))$, we have 
$$[X_{\alpha\gamma}][X^{(\delta\beta)^\vee}]=  [X_{\beta\gamma}][X^{(\delta\alpha)^\vee}] =  [X_{\alpha\delta}][X^{(\gamma\beta)^\vee}] = [X_{\beta\delta}][X^{(\gamma\alpha)^\vee}].$$

\end{enumerate}


\end{lemma}

\begin{proof}[Proof idea] Applying \Cref{lemma:directsum} and a simple geometric argument gives us that 
    {\parfillskip = 0pt \par} \noindent
    $[X_{\alpha\gamma}][X^{(\delta\beta)^\vee}] = \Omega_*([X_\alpha][X^{\beta^\vee}]\otimes [X_\gamma][X^{\delta^\vee}])= \Omega_*((J_a,I_c)\cdot ([X_\beta][X^{\alpha^\vee}]\otimes [X_\gamma][X^{\delta^\vee}])) = \Phi_a\cdot \Omega_*([X_\beta][X^{\alpha^\vee}]\otimes [X_\gamma][X^{\delta^\vee}]) = \Phi_a\cdot ([X_{\beta\gamma}][X^{(\delta\alpha)^\vee}])$. A similar argument gives the other two equalities in (i). Then (ii) follows from the fact that, in ordinary cohomology $H^*(\Gr(a_1+c_1;\C^{a+c})$, $\Phi_a$ and $\Phi_c$ act trivially on the classes.
\end{proof}




\section{Proofs of Theorems}



\subsection{Puzzles with split symmetry and trapezoidal puzzles}

\subsubsection{Proof of \texorpdfstring{\Cref{thms:commutativeproperties}}{Theorem 1.1}(a) (puzzles with split symmetry)}




\begin{proof}
\textbf{Geometric Proof.} The claim is equivalent to 
\begin{multline*}
    \underset{\Gr_{a+c}}{\int} [X_{\alpha\gamma}][X^{(\delta\beta)^\vee}][X_\nu] = \underset{\Gr_{a+c}}{\int} [X_{\beta\gamma}][X^{(\delta\alpha)^\vee}][X_\nu]\\
    = \underset{\Gr_{a+c}}{\int} [X_{\alpha\delta}][X^{(\gamma\beta)^\vee}][X_\nu] = \underset{\Gr_{a+c}}{\int} [X_{\beta\delta}][X^{(\gamma\alpha)^\vee}][X_\nu],
    \end{multline*}
which follows immediately from \Cref{lemma:commutesplit}.

\textbf{Proof by the commutative property of triangular puzzles.} 
     
    In any $\inlinetri_{\alpha\gamma,\nu,\delta\beta}$-puzzle, due to the symmetries in string content, there can be no $\tenstr$s appearing along the red line shown in \Cref{fig:splitsymmetrypuzzle}.
    So we can divide the puzzle into a triangular puzzle and a trapezoidal puzzle. Then we can sum over choices of label $\lambda$ on the red line and obtain
\begin{gather*}
\#\{\inlinetri_{\alpha\gamma,\nu,\delta\beta} \text{-puzzles}\} = \sum_\lambda \left(\#\{\inlinetri_{\alpha,\lambda,\beta} \text{-puzzles}\}\right) \left(\#\{\inlinetrapside_{\lambda^\vee,\gamma,\nu,\delta} \text{-puzzles}\}\right).
\end{gather*} 
Since $\#\{\inlinetri_{\alpha,\lambda,\beta} \text{-puzzles}\}=\#\{\inlinetri_{\beta,\lambda,\alpha} \text{-puzzles}\}$ by the commutative property of triangular puzzles and three-fold rotational symmetry, this implies that $\alpha$ and $\beta$ commute.


To prove that $\gamma$ and $\delta$ commute, we first apply the commutative property of triangular puzzles to the NW and NE sides of the outer boundary to get
$\#\{\inlinetri_{\alpha\gamma,\nu,\delta\beta} \text{-puzzles}\}= \#\{\inlinetri_{\nu,\alpha\gamma,\delta\beta} \text{-puzzles}\}$, which leads to the picture in \Cref{fig:splitsymmetrypuzzlecommuted}. Then we follow a similar argument as the one used to commute $\alpha$ and $\beta$ above.
\end{proof}



\begin{figure}[ht]
    \centering
    \begin{subfigure}[t]{0.47\textwidth}
    \centering
     \scalebox{.8}{\input{splitsymmetrypuzzle}}
    \caption{$\alpha$ and $\beta$ commute via the commutative property of triangular puzzles.
    %the basic commutative property of triangular puzzles (plus rotational symmetry) applied to the green triangular subpuzzle.
    }
    \label{fig:splitsymmetrypuzzle}   
    \end{subfigure}
    ~
    \begin{subfigure}[t]{0.47\textwidth}
    \centering
    \scalebox{.8}{\input{splitsymmetrypuzzlecommuted}}
    \caption{$\gamma$ and $\delta$ commute via the commutative property of triangular puzzles.
    %the basic commutative property of triangular puzzles (plus rotational symmetry) applied to the green triangular subpuzzle.
    }
    \label{fig:splitsymmetrypuzzlecommuted}
    \end{subfigure}
    \caption{}
    \label{fig:splitsymmetrypuzzles}
\end{figure}





\subsubsection{Proof of \texorpdfstring{\Cref{thms:commutativeproperties}}{Theorem 1.1}(b) (trapezoidal puzzles)}


\begin{proof}
    \textbf{Proof by the commutative property of puzzles with split symmetry.} After completing to a triangle as in \Cref{fig:trapezoidcomplete}, we can immediately apply \Cref{thms:commutativeproperties}(a). 
    %to commute $\gamma$ and $\delta$.
\end{proof}








\subsection{Parallelogram-shaped puzzles}


\subsubsection{Proof of \texorpdfstring{\Cref{thms:commutativeproperties}}{Theorem 1.1}(c) (parallelogram-shaped puzzles)}


\begin{proof}

For all of the proofs below, we begin by completing to a triangle as in \Cref{fig:parallelogramcomplete}. 

    \textbf{Geometric proof.} The claim is equivalent to 
    \begin{multline*}
        \underset{\Gr_{a+c}}{\int} [X_{\sort(\alpha)\gamma}][X_{\beta\sort(\delta)}][X^{(\delta\alpha)^\vee}] = \underset{\Gr_{a+c}}{\int} [X_{\sort(\alpha)\gamma}][X_{\alpha\sort(\delta)}][X^{(\delta\beta)^\vee}] \\
        = \underset{\Gr_{a+c}}{\int} [X_{\sort(\alpha)\delta}][X_{\beta\sort(\delta)}][X^{(\gamma\alpha)^\vee}] = \underset{\Gr_{a+c}}{\int} [X_{\sort(\alpha)\delta}][X_{\alpha\sort(\delta)}][X^{(\gamma\beta)^\vee}],
    \end{multline*}
    which can be easily obtained through applications of \Cref{lemma:commutesplit}.

    \textbf{Proof by the commutative property of triangular puzzles.} We first apply the commutative property of triangular puzzles to the NW and NE sides of the outer boundary to get
$\#\{\inlinetri_{\beta\sort(\delta),\sort(\alpha)\gamma,\delta\alpha} \text{-puzzles}\}= \#\{\inlinetri_{\nu,\alpha\gamma,\delta\beta} \text{-puzzles}\}.$ From there, we can immediately commute the pairs $\alpha,\beta$ and $\gamma,\delta$ in a manner similar to that in \Cref{fig:splitsymmetrypuzzles}.

     \textbf{Proof by the commutative property of puzzles with split symmetry.} This property (given in \Cref{thms:commutativeproperties}(a)) combined with the three-fold rotational symmetry of puzzles proves the statement.
\end{proof}




\subsubsection{\texorpdfstring{Proof of \Cref{thms:commutativeproperties}(d) (symmetric rhombus-shaped puzzles)}{Proof of Theorem 1.1(d) (symmetric rhombus-shaped puzzles)}}


\begin{proof}

    \textbf{Geometric proof.} A geometric proof is given in \cite{anderson2024} but is omitted here.

    \textbf{Proof by the commutative properties of triangular and parallelogram-shaped puzzles.} This special case of a parallelogram-shaped puzzle can be seen as two triangular puzzles glued together. We apply \Cref{thms:commutativeproperties}(c) along with the commutative property of triangular puzzles to obtain all permutations of the rhombus's boundary labels.
\end{proof}



\subsection{Hexagonal puzzles}

\texorpdfstring{Proof of Theorems \Cref{thms:commutativeproperties}(f) (hexagonal puzzles with opposite sides symmetry)}{Proof of Theorems Theorem 1.1(f) (hexagonal puzzles with opposite sides symmetry)}

\subsubsection{\texorpdfstring{Proof of Theorems \Cref{thms:commutativeproperties}(f) (hexagonal puzzles with opposite sides symmetry)}{Proof of Theorems Theorem 1.1(f) (hexagonal puzzles with opposite sides symmetry)}}

\begin{proof}
\textbf{Geometric proof.} A geometric proof is given in \cite{anderson2024} but is omitted here.
    
    
    \textbf{Proof by the commutative property of parallelogram-shaped puzzles.} We can complete a hexagonal puzzle to a parallelogram-shaped puzzle to obtain a bijection $\{\inlinehex_{\alpha,\beta,\gamma,\delta,\epsilon,\zeta} \text{-puzzles}\} \leftrightarrow \{\inlinepar_{\sort(\zeta)\alpha,\beta\gamma,\sort(\gamma)\delta,\epsilon\zeta} \text{-puzzles}\}.$ We can then use \Cref{thms:commutativeproperties}(c) to swap the SW and NE labels, i.e. $\sort(\zeta)\alpha$ and $\sort(\gamma)\delta$, which effectively swaps $\alpha$ and $\delta$, as $\sort(\gamma)=\sort(\zeta)$. Combining this with three-fold rotational symmetry yields the other swaps.
\end{proof}



\subsubsection{\texorpdfstring{Proof of \Cref{thms:commutativeproperties}(g) (hexagonal puzzles with two-way symmetry)}{Proof of Theorem 1.1(g) (hexagonal puzzles with two-way symmetry)}}




\begin{proof}
    \textbf{Geometric proof.} 
    A geometric proof is given in \cite{anderson2024} but is omitted here.
  

    \textbf{Proof by the commutative property of trapezoidal puzzles}: A hexagonal puzzle with this type of symmetry can be seen as two trapezoidal puzzles glued together. Thus, the claim follows easily from \Cref{thms:commutativeproperties}(b).
\end{proof}

\subsubsection{\texorpdfstring{Proof of \Cref{thms:commutativeproperties}(e),(h),(i)}{Proof of Theorem 1.1(e),(h),(i)}}
\begin{proof} These statements can be derived from \Cref{thms:commutativeproperties}(f) and \Cref{thms:commutativeproperties}(g), as they are just more specialized cases of those types of symmetries.
\end{proof}












\subsection{Commutative property of parallelogram-shaped equivariant puzzles}
\label{section:commutativepropertyofeqvtparallelogram}


The equivariant structure constants in the theorem below are the ones associated to parallelogram-shaped equivariant puzzles, after completing to a triangle as in \Cref{fig:parallelogramcomplete}.

\begin{theorem}[Commutative Property of Structure Constants Associated to Parallelogram-Shaped Equivariant Puzzles]
    \label{thm:eqvtparallelogram}

   Let $\Phi_a$ and $\Phi_c$ be defined as in \Cref{lemma:commutesplit}.
Let $\alpha,\beta\in\binom{[a]}{a_1}$ and $\gamma,\delta\in\binom{[c]}{c_1}$. Then in $H^*_T(\Gr(a_1+c_1;\C^{a+c}))$, we have 
\begin{multline*}(c_T)_{\sort(\alpha)\gamma,\beta\sort(\delta)}^{(\delta\alpha)^\vee}= \Phi_a\cdot (c_T)_{\sort(\beta)\gamma,\alpha\sort(\delta)}^{(\delta\beta)^\vee} \\
= \Phi_c \cdot (c_T)_{\sort(\alpha)\delta,\beta\sort(\gamma)}^{(\gamma\alpha)^\vee} = \Phi_c\cdot\Phi_a\cdot (c_T)_{\sort(\beta)\delta,\alpha\sort(\gamma)}^{(\gamma\beta)^\vee}.\end{multline*}
      
In other words commuting the pair $\alpha, \beta$ reverses the $y_1,\ldots,y_a$, and commuting the pair $\gamma,\delta$ reverses the $y_{a+1},\ldots,y_{a+c}$, in the structure constant.
\end{theorem} 

\begin{proof}[Proof idea]
    %In \cite{anderson2024}, we show that
    It is not difficult to show that $\Phi_a\cdot X_{\sort(\alpha)\gamma}= X_{\sort(\alpha)\gamma}$, and this along with \Cref{lemma:commutesplit} gives 
    \begin{multline*}
    \Phi_a\cdot (c_T)_{\sort(\beta)\gamma,\alpha\sort(\delta)}^{(\delta\beta)^\vee}  = \underset{\Gr_{a+c}}{\int} \left(\Phi_a\cdot[X_{\sort(\alpha)\gamma}]\right) \left(\Phi_a\cdot([X_{\alpha\sort(\delta)}][X^{(\delta\beta)^\vee}])\right)\\
    = \underset{\Gr_{a+c}}{\int} [X_{\sort(\alpha)\gamma}] [X_{\beta\sort(\delta)}][X^{(\delta\alpha)^\vee}] =(c_T)_{\sort(\alpha)\gamma,\beta\sort(\delta)}^{(\delta\alpha)^\vee}.
\end{multline*}
A similar argument proves the other two equalities in the claim.
\end{proof}


\begin{example}
\normalfont   See \Cref{fig:eqvtparexample}.
\end{example}


\begin{figure}[ht]
    \centering
    \begin{subfigure}[t]{0.47\textwidth}
    \centering
     \input{eqvtexample_v1_row}
    \caption{The set of all $\inlinepar_{\zerostr\zerostr\onestr\onestr,\zerostr\onestr\zerostr,\onestr\onestr\zerostr\zerostr,\zerostr\zerostr\onestr}$-puzzles. Sum of the puzzle weights: $(y_5-y_1)(y_7-y_1)(y_7-y_2)+(y_6-y_2)(y_7-y_1)(y_7-y_2)$
    }
    \end{subfigure}
    ~
    \begin{subfigure}[t]{0.47\textwidth}
    \centering
    \input{eqvtexample_v2_row}
    \caption{The set of all $\inlinepar_{\onestr\onestr\zerostr\zerostr,\zerostr\onestr\zerostr,\zerostr\zerostr\onestr\onestr,\zerostr\zerostr\onestr}$-puzzles. Sum of the puzzle weights: $(y_6-y_4)(y_7-y_3)(y_7-y_4)+(y_5-y_3)(y_7-y_3)(y_7-y_4)$
    }
    \end{subfigure}
    \caption{The effect of commuting the labels on the SW and NE sides is reversing the $y_1,y_2,y_3,y_4$ in the structure constant.}
    \label{fig:eqvtparexample}
\end{figure}

\section{Further Questions}


So far, we relate polygonal puzzles to special classes of triangular puzzles and their associated structure constants, but it is unknown if polygonal puzzles hold a more intrinsic meaning 
%in geometry, combinatorics, or representation theory.
geometrically, combinatorially, or representation-theoretically. 
It would be interesting to explore this question and to find any illuminating manifestations or proofs of these commutative properties in the various contexts where LR numbers arise.


\acknowledgements{We thank Allen Knutson, Pasha Pylyavskyy, and Paul Zinn-Justin for their support.}

%% if you use biblatex then this generates the bibliography
%% if you use some other method then remove this and do it your own way
\printbibliography



\end{document}
