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\newcommand\comaj{{\mathop\mathrm{comaj}}}

\newcommand\martial{{\Large \mars\!\!}}
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\usepackage{lipsum,wasysym, mathtools}

%% define your title in the usual way
\title{The commutant of divided difference operators, Klyachko's genus,
  and the $\comaj$ statistic}

%% define your authors in the usual way
%% use \addressmark{1}, \addressmark{2} etc for the institutions, and use \thanks{} for contact details
\author{Christian Gaetz\thanks{\href{mailto:gaetz@berkeley.edu}{gaetz@berkeley.edu}}\addressmark{1}, 
  Rebecca Goldin\thanks{\href{mailto:rgoldin@gmu.edu}{rgoldin@gmu.edu}. RG was partially supported by NSF grant DMS--2152312.}\addressmark{2},
\and Allen Knutson\thanks{\href{mailto:allenk@math.cornell.edu}{allenk@math.cornell.edu}. AK was partially supported by NSF grant DMS--2246959.}\addressmark{3}
}

%% then use \addressmark to match authors to institutions here
\address{
\addressmark{1} Department of Mathematics, University of California, Berkeley \\ 
\addressmark{2} Department of Mathematical Sciences, George Mason University \\
\addressmark{3} Department of Mathematics, Cornell University}

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\received{\today}

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%% put your English abstract here, or comment this out if you don't have one yet
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\abstract{In \cite{ZachEtAl,Nenashev,AnnaOliver} are studied certain
  operators on polynomials and power series that commute with all
  divided difference operators $\partial_i$. We introduce a second set
  of ``martial'' operators $\martial_i$ that generate the full commutant,
  and show how a Hopf-algebraic approach naturally reproduces the operators
  $\xi^\nu$ from \cite{Nenashev}. We then pause to study Klyachko's homomorphism
  $H^*(Fl(n)) \to H^*($the permutahedral toric variety$)$, and extract the
  part of it relevant to Schubert calculus, the ``affine-linear genus''.
  This genus is then re-obtained using Leibniz combinations of the
  $\martial_i$. We use Nadeau-Tewari's $q$-analogue of Klyachko's genus
  to study the equidistribution of $\ell$ and $\comaj$
  on $[n]\choose k$, generalizing known results on $S_n$.
  }

%% 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{divided difference operators, Schubert calculus, comaj statistic}

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\usepackage[backend=bibtex]{biblatex}
\addbibresource{martials.bib}

\begin{document}

\maketitle
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% \tableofcontents



\section{The martial operators $\martial_\pi$}


\subsection{The ring of Schubert symbols}

Given a Dynkin diagram $D$ with Weyl group $W(D)$, define the
\defn{ring of Schubert symbols} $H(D)$ as the cohomology ring of
the associated (possibly infinite-dimensional) flag variety,
with the usual Schubert basis $\{\calS_w\colon w\in W(D)\}$. 
The Dynkin diagrams that will interest us are primarily the
semi-infinite $A_{\ZZ_+}$ and the biinfinite $A_\ZZ$.
% While ``$H(D)$'' was
% chosen to suggest cohomology (of a flag variety), we didn't want to
% deal with infinite-dimensional geometry, and take a purely algebraic approach.
In these type $A$ cases $W(D)$ is the group of finite permutations
of $\ZZ_+$ or of $\ZZ$. An important difference between the two is
that $H(A_{\ZZ_+})$ is generated by $\{\calS_{r_i}\colon i\in \ZZ_+\}$, where $r_i$ is a simple transposition,
so the multiplication is entirely determined by Monk's rule, whereas $H(A_{\ZZ})$
requires additional generators $\{\calS_{r_1r_2\cdots r_k}\}$ and determining
its multiplication involves also the flag Pieri rule.
With all that in mind we largely abandon the geometry and work with these
rings symbolically.
\junk{
  and the (Chevalley-)Monk formula is % just the Monk formula
  $\calS_\pi\, \calS_i = \sum \{\calS_{\pi'}\colon \pi' \gtrdot \pi,\ \pi' = \pi \circ (a\ b),\ a \leq i<b\}$.
}

For each vertex $\alpha$ of $D$ hence generator $r_\alpha \in W(D)$,
we have an operator $\partial_\alpha \actson H(D)$,
pronounced ``partial $\alpha$'':
$$ \partial_\alpha \,\calS_\pi :=
\begin{cases}
  \calS_{ \pi r_\alpha} &\text{if }  \pi r_\alpha < \pi \\
  0 &\text{if } \pi  r_\alpha > \pi
\end{cases}
$$
from which we can well-define $\partial_{\pi}$ for any $\pi \in W(D)$ using products.

\begin{thm}[Lascoux-Sch\"utzenberger]\label{thm:LS}
  There is an isomorphism $H(A_{\ZZ_+}) \to \ZZ[x_1,x_2,\ldots]$
  taking the Schubert symbol $\calS_\pi$ to its ``Schubert polynomial''
  $S_\pi(x_1,x_2,\ldots)$. On the target ring, $\partial_\alpha$ acts
  by Newton's divided difference operation.
\end{thm}

Call a ring homomorphism from $H(D)$ to some other ring a
\defn{genus},\footnote{This terminology is stolen from the study of
  various cobordism rings of a point, e.g. the ``Hirzebruch genus''
  and ``Witten genus'' are ring homomorphisms to $\ZZ$.}
making the above isomorphism the \defn{Lascoux-Sch\"utzenberger genus}.

It was observed in \cite{ZachEtAl,AnnaOliver} that the operator
$\nabla := \sum_i \frac{d}{dx_i}$ on the target ring has two nice properties:
 it commutes with each $\partial_i$, and
 its application to any Schubert polynomial is a positive combination
  of Schubert polynomials.
Our goal in this section is to characterize operations of the
first type, with an eye toward the second.
To study this commutant it will be handy to work with algebra actions.

\subsection{Two commuting actions of the nil Hecke algebra}
\label{ssec:twoactions}

Let $Nil(D)$ denote formal (potentially infinite) linear combinations of
elements $\{d_\pi \colon \pi \in W(D)\}$, with a multiplication defined by
$ d_\pi d_\rho :=
\begin{cases}
  d_{\pi \rho} &\text{if }\ell(\pi \rho) = \ell(\pi) + \ell(\rho) \\
  0 &\text{if }\ell(\pi \rho) < \ell(\pi) + \ell(\rho).
\end{cases}
$
This multiplication extends to infinite sums in a well-defined way, insofar
as any $w\in W(D)$ has only finitely many length-additive
factorizations. Slightly abusing\footnote{One ordinarily considers only
  finite linear combinations, but we have need of certain infinite ones,
  and this simplifies the statement of Theorem \ref{thm:commutant}.}
terminology, we call this $Nil(D)$ the \defn{nil Hecke algebra}.
The association $d_\pi \mapsto \partial_\pi$ gives an action
of the opposite algebra $Nil(D)^{op}$ on $H(D)$; the infinitude
of the sums in $Nil(D)$ is again not problematic, because $H(D)$'s elements
are finite sums of Schubert symbols.

Define $\martial_\alpha$ (pronounced ``martial $\alpha$'') by
$$ \martial_\alpha \calS_\pi :=
\begin{cases}
  \calS_{r_\alpha \pi} &\text{if } r_\alpha \pi < \pi \\
  0 &\text{if } r_\alpha \pi > \pi 
\end{cases}
$$ \hskip -.05in % why the hell is this necessary here?
We can well-define $\martial_{\prod Q} := \prod_{q\in Q} \martial_q$
for each reduced word $Q$.
  
\begin{thm}\label{thm:commutant}
  The map $d_\pi \mapsto \martial_\pi$ defines an action of $Nil(D)$
  on $H(D)$ (unspoiled by the potential infinitude), commuting with
  the $Nil(D)^{op}$-action by the operators $\partial_\alpha$.
  Conversely, each operator on $H(D)$ that commutes with all operators
  $\partial_\alpha$ arises as the action of a unique element of $Nil(D)$.

  In short, $Nil(D)$ and $Nil(D)^{op}$ are one another's commutants
  in their actions on $H(D)$.
\end{thm}

This then characterizes the operators that commute with all $\partial_i$;
we don't know any significance of the resulting algebra again being $Nil(D)$.   

\begin{proof}
  For $h \in H(D)$, let $\int h$ denote the coefficient of $\calS_e$ in $h$.
  Each $h = \sum_\pi h_\pi \calS_\pi$ is determined by the values
  $$ \int (\partial_\rho h) = \int \sum_\pi h_\pi \partial_\rho \calS_\pi
  = \sum_\pi h_\pi \int \partial_\rho \calS_\pi
  = \sum_\pi h_\pi \delta_{\pi,\rho} = h_\rho 
$$ \hskip -.05in % why the hell is this necessary here?
  We'll make use of the easy fact
  $ \int \martial_\pi \,\partial_\rho \,\calS_\sigma =
  \begin{cases}
    1 &\text{if $\sigma= \pi^{-1} \rho$ and }\ell(\sigma)=\ell(\pi)+\ell(\rho)\\
    0 &\text{otherwise.}
  \end{cases}
  $
  
  Now let $C$ be an operator on $H(D)$ commuting with all $\partial_\pi$.
  For each $\pi\in W(D)$, let $c_\pi := \int C(\calS_{\pi^{-1}})$.
  We then confirm $C = \sum_\pi c_\pi \martial_\pi$ using the
  determination above:
  \begin{eqnarray*}
    \int \partial_\rho C(\calS_\sigma)
    &=& \int C(\partial_\rho \calS_\sigma)
        = \int C\left(\calS_{\sigma \rho^{-1}}\, 
        \left[\ell(\sigma \rho^{-1})=\ell(\sigma)-\ell(\rho)\right] \right) \\
    &=& \left[\ell(\sigma \rho^{-1})=\ell(\sigma)-\ell(\rho)\right]\ 
        c_{\rho \sigma^{-1}} \\
    \int  \partial_\rho \left( \sum_\pi c_\pi \martial_\pi\right) (\calS_\sigma)
    &=&  \sum_\pi c_\pi \int \martial_\pi \,\partial_\rho \,\calS_\sigma 
    \ \ =\ \ \sum_\pi c_\pi \, \left[\sigma 
= \pi^{-1}\rho\right]\, \big[\ell(\sigma)=\ell(\pi)+\ell(\rho)\big] \\
    &=& c_{\rho \sigma^{-1}} \left[\ell(\rho\sigma^{-1})=\ell(\sigma)-\ell(\rho)\right]
%        \qquad\qquad\qquad\qquad\qquad\qquad\quad \qedhere
  \end{eqnarray*}
  Here $[P]=1$ if $P$ is true, $[P]=0$ if $P$ is false, for a statement $P$.
\end{proof}

{\em Example.} The action of $\nabla := \sum_i \frac{d}{dx_i}$ on polynomials,
pulled back to an action on $H(A_{\ZZ_+})$, is given by the operator
$\sum_{n \in \NN_+} n \martial_n$.
What is particularly special about $\nabla$ is that it is
a differential (i.e. satisfies the Leibniz rule), and is of degree $-1$.

\begin{thm}\label{thm:Leibniz}
  Let $\sum_\alpha c_\alpha \martial_\alpha \in Nil(D)$ be an operator of
  degree $-1$. If it is a differential (and $D$ is simply-laced,
  for convenience) then each $c_\alpha$ is $\frac 1 2 \sum_\beta c_\beta$
  where the $\beta$s are $\alpha$'s neighbors in $D$.

  In particular if $D$ is of finite type $ADE$, the only system
  of coefficients $(c_\alpha)$ is zero. If $D=A_{\ZZ_+}$, the only options
  are multiples of $c_i \equiv i$. If $D=A_\ZZ$, the space of such systems
  is two-dimensional, spanned by $c_i \equiv i$ and $c_i \equiv 1$.
\end{thm}

Hence the $\nabla$ discovered in \cite{ZachEtAl} in the $A_{\ZZ_+}$ case was
the only such operator available. In \cite{Nenashev} it is explained
that $\xi=\sum_{i \in \ZZ} \martial_i$ is special to the back-stable situation of $A_\ZZ$; here
we see that it is the only new option. (The result
\cite[Theorem 6]{Nenashev} is very similar.)

\begin{proof}[Proof sketch]
  The proof amounts to applying $\sum_\alpha c_\alpha \martial_\alpha$ to
  $(\calS_{r_\alpha})^2 = \sum_\beta \calS_{r_\beta r_\alpha}$
  (computed using the Chevalley-Monk rule).
\end{proof}

% ... sufficient in the type $A$ cases, maybe not elsewhere?

\section{Not-quite-Hopf algebras and Nenashev operators}

\subsection{The dual algebras}

Define a pairing $Nil(D) \tensor_\ZZ H(D) \rightarrow \ZZ$ by
$$
 p \tensor s \mapsto \text{coefficient of $\calS_e$ in }p(s)
$$
%\begin{eqnarray*}
%  Nil(D) \tensor_\ZZ H(D) &\to& \ZZ \\
%  p \tensor s &\mapsto& \text{coefficient of $\calS_e$ in }p(s)  
%\end{eqnarray*}
and from there a map $Nil(D) \to H(D)^* := Hom_\ZZ(H(D),\ZZ)$.

The following is well-known to the experts, if not usually expressed
exactly this way (see e.g. \cite{BerensteinRichmond},
\cite[\S7.2]{LLS}). 

\begin{thm}\label{thm:Hopf}
  This map $Nil(D) \to H(D)^*$ is an isomorphism. Unfortunately the induced
  comultiplication $H(D) \to H(D)\tensor H(D)$ is not a ring homomorphism
  (example below), so the two are not thereby dual Hopf algebras.
  (There is an alternative statement explored in \cite{LXZ}.)
\end{thm}

\newcommand\BS{BS}


There is an analogue of Theorem \ref{thm:LS} for $H(A_\ZZ)$, taking each
$\calS_\pi$ to its
\defn{back-stable Schubert function} $\BS_\pi$ invented by the third author
(and independently by Buch and by Lee), which were studied in
\cite{LLS,Nenashev}. Define a \defn{back-stable function}
$p \in \ZZ[[\ldots,x_{-1},x_0,x_1,x_2,\ldots]]$ to be a power series
\begin{itemize}
\item of finite degree, such that
\item $p$ depends only on the variables $\{x_k, k<N\}$ for some $N\gg 0$, and
\item for some $M\ll 0$, p is symmetric in the variables $\{x_i, i\leq M\}$.
\end{itemize}
One way (as appears in \cite{Nenashev}) to think of the ring of
back-stable functions is as the image of the injection
\begin{eqnarray*}
  Symm \tensor_\ZZ \ZZ[\ldots,x_{-1},x_0,x_1,\ldots] &\to&
\ZZ[[\ldots,x_0,\ldots]] / \langle 
        \text{elementary symmetric functions}\rangle \\
p \tensor q &\mapsto& p(\ldots,x_{-2},x_{-1},x_0) \, q
\end{eqnarray*}

For $\pi \in W(A_\ZZ)$ considered as a finite permutation of $\ZZ$,
%and $\pi[N] \coloneqq shift_N(i) \equiv i+N$, observe for $N\gg 0$ that
and $shift_N(i) := i+N$, observe for $N\gg 0$ that $\pi[N] \coloneqq shift_N(i)\circ \pi\circ shift_{-N}(i)$
%$\pi[N]$
 is a permutation of $\ZZ$ that
leaves $-\NN$ in place, and thus has a well-defined Schubert
polynomial. Define the \defn{back-stable Schubert function}
$$ \BS_\pi := \lim_{N\to \infty} 
S_{\pi[N]}(x_{1-N},x_{2-N},\ldots) 
$$
% \hskip -.05in % why the hell is this necessary here?
where the limit is computed coefficient-wise (note that any single coefficient settles down to a constant value for all large enough $N$). 

\begin{thm}\cite[Theorem 3.5]{LLS}
  \label{thm:backstable}
  The back-stable Schubert functions lie in, and are a $\ZZ$-basis of,
  the ring of back-stable functions. 
\end{thm}

In this co\"ordinatization we can compute the comultiplication on $H(A_\ZZ)$
and bound its failure to be a ring homomorphism. Transposing the
multiplication from \S\ref{ssec:twoactions} of $d_\pi d_\rho$, we obtain
$\Delta(\BS_\sigma) = \sum \{BS_\pi \tensor BS_\rho\colon
\sigma = \pi \rho,\ \ell(\sigma) = \ell(\pi)+\ell(\rho)\}$.
Then, alas,
\newcommand\ttensor{{\!\!\>\otimes\!\>\!}}  % tightened up a little bit
\begin{eqnarray*}
  \Delta(\BS_{r_2}^2) =   \Delta(\BS_{r_1r_2} + \BS_{r_3r_2})
                          &=& (\BS_{r_1r_2}\ttensor 1)
                          + (\BS_{r_1}\ttensor \BS_{r_2})
                          + (1\ttensor \BS_{r_1r_2}) \\
                          &&+ (\BS_{r_3r_2}\ttensor 1)
                          + (\BS_{r_3}\ttensor \BS_{r_2})
                             + (1\ttensor \BS_{r_3r_2})\\
\neq\Delta(\BS_{r_2}) \Delta(\BS_{r_2}) = (\BS_{r_2}\ttensor 1 + 1\ttensor\BS_{r_2})^2
                          &=& (\BS_{r_1r_2}\ttensor 1)
                          + (\BS_{r_3r_2}\ttensor 1)
                          + (\BS_{r_2}\ttensor \BS_{r_2}) \\
                          &&+ (\BS_{r_2}\ttensor \BS_{r_2})
                          + (1 \ttensor \BS_{r_1r_2})
                             + (1\ttensor \BS_{r_3r_2} )
\end{eqnarray*}
Luckily $\Delta(\BS_\pi \BS_{\rho[N]})
= \Delta(\BS_{\pi \circ (\rho[N])})
= \Delta(\BS_{\pi}) \Delta(\BS_{\rho[N]})$
for $N\gg 0$. Call this property \defn{``separated Hopfness''}, to be
used below.

\subsection{The Fomin-Greene--Nenashev operators $\xi^\nu$}

With these identifications, and the self-duality of the Hopf algebra
$Symm$ of symmetric functions, we can interpret some results
of Nenashev \cite{Nenashev}:
$$
\begin{matrix}
  H(A_\ZZ) &\xrightarrow{\sim}& \{\text{back-stable functions}\}
  &\xleftarrow{\sim}& Symm \tensor_\ZZ \ZZ[\ldots,x_{-1},x_0,x_1,\ldots]
  &\onto& Symm \\ \\
  Nil(A_\ZZ) & & & \from & & & Symm
\end{matrix}
$$

The map $\onto$ is the \defn{Stanley genus}:
it takes $\calS_\pi$ to its \defn{Stanley symmetric function}
$St_\pi = \sum_\lambda a_\pi^\lambda\, Schur_\lambda$. The lower map, its transpose,
takes $Schur_\lambda$ to $\sum_\pi a_\pi^\lambda\, d_\pi$. If we let this
operator act on $H(A_\ZZ)$ under the $d_\pi \mapsto \martial_\pi$ action, we get
the \defn{Fomin-Greene--Nenashev operator}
$\xi^\lambda := \sum_\pi a_\pi^\lambda \,\martial_\pi$ \cite{FominGreene,Nenashev}.
(See also the $j_\lambda$ operators in the ``Peterson subalgebra''
defined in \cite[\S9.3]{LLS}, which are a double version of the $\xi^\lambda$.)


\newcommand\mm{{\mathfrak m}}
Let $\mm$ denote the kernel of the map $H(A_\ZZ) \onto Symm$.
Using the separated Hopfness and the fact that
$\BS_{\pi[N]}-BS_\pi \in \mm$, one shows that each
$\Delta(pq)-\Delta(p)\Delta(q)$
(which serves as a measure of non-Hopfness)
lies in $\mm \tensor H(A_\ZZ) + H(A_\ZZ) \tensor \mm$. Hence the map 
$H(A_\ZZ)\to Symm$ factors through a map of Hopf algebras. Dually, the
transpose map is a Hopf map to a Hopf sub-bialgebra of $Nil(A_\ZZ)$.
In particular this Hopf map explains Nenashev's
formul\ae\ \cite[\S 4.4]{Nenashev}
$$ \xi^\lambda \xi^\mu = \sum_\nu c_{\lambda\mu}^\nu\, \xi^\nu
\qquad\qquad \xi^\nu(pq) = \sum_{\lambda,\mu} c_{\lambda\mu}^\nu \, \xi^\lambda(p)
\,\xi^\mu(q) $$

\subsection{Interlude (not used elsewhere):
  topological origin of the $\{BS_\pi\}$}

\junk{
  While we won't need it for our other results, we present an origin story
  for the back-stable Schubert functions. }

The stability property underlying Lascoux-Sch\"utzenberger's definition
of Schubert polynomials is the fact that each $S_\pi \in H^*(Fl(n))$
is the pullback $\iota_{n+1}^*(S_{\pi\oplus 1})$ along a map
$\iota_{n+1}: Fl(n) \into Fl(n+1)$ taking $(E_\bullet)$ to
$(F_\bullet\colon F_{i\leq n} = E_i\oplus 0,\ F_{n+1} = E_n \oplus \CC)$.
Chaining these together, one builds an element of the inverse limit
of the cohomology rings, a ring
$ \ZZ[[x_1,x_2,\ldots]]/\langle$elementary symmetric functions $e_i\rangle$.
It was then Lascoux-Sch\"utzenberger's pleasant surprise that these
``inverse limit Schubert classes'' lie in (and exactly span) the image of the
injective ring homomorphism $\ZZ[x_1,x_2,\ldots]$ into this algebra.

This admits of a parallel story, based on a different map
$\iota_{1+2n+1}: Fl(2n) \into Fl(2n+2)$ taking $(E_\bullet)$ to
$(F_\bullet\colon F_{i\in [1,2n+1]} = \CC \oplus E_{i-1}\oplus 0,\
F_{2n+2} = \CC \oplus \CC^{2n} \oplus \CC)$. Now, in order to achieve a
coherent labeling (as $n$ varies)
we index the classes in $H^*(Fl(2n))$ using permutations
of $[1-n, n]$ rather than of $[1,2n]$. Once again the inverse limit is
a power series ring modulo elementary symmetrics, but it is {\em no
  longer true} that the inverse limit Schubert classes are representable
by polynomials; rather, they can be represented by back-stable functions.
(And again, they form a basis thereof.)

One advantage of $\iota_{1+2n+1}$ is that it is equivariant w.r.t. the
{\em duality} endomorphism of $Fl(2n)$, which takes $(E_\bullet)$ to
$(E_\bullet^\perp)$,
defined w.r.t. the symplectic form pairing co\"ordinates $i$ and $1-i$, for
$i\in [1,n]$. On the level of classes, this
takes $BS_\pi \mapsto BS_{w_0 \pi w_0}$
where $w_0(i) := 1-i$. On the level of back-stable functions,
it takes $x_i \mapsto -x_{1-i}$,\ $e_i(x_{\leq 0}) \mapsto e_i(x_{\leq 0})$.

Since this duality respects Schubert classes and the alphabet $(x_i)$,
it takes Monk's rules to Monk's rules. In particular it turns the
transition formula (a specific Monk's rule)
$$ BS_\pi = x_i BS_{\pi'} + \sum_{\text{certain }\pi''} BS_{\pi''}
\qquad\qquad\text{into}\qquad\qquad
BS_\rho = -x_j BS_{\rho'} + \sum_{\text{certain }\rho''} BS_{\rho''} $$
\hskip -.05in
which implies (unstably) the cotransition formula
$x_j \calS_{\rho'} = -\cancel{\calS_\rho}
+ \sum_{\text{certain }\rho''} \calS_{\rho''} $
of \cite{K}.


\section{Relation to Klyachko's genus}

\subsection{Klyachko's ideal and its prime factors}

Let $T \leq GL_n(\CC)$ denote the group of diagonal matrices, and
$TV_{perm} \subseteq Fl(n)$ be the \defn{permutahedral toric variety}
obtained as the closure of a generic $T$-orbit on the flag manifold $Fl(n)$.
This subvariety arises as a Hessenberg variety (see e.g. \cite{abe2020survey})
and is of key importance in \cite{Huh,NadeauTewari}. 

The inclusion $\iota\colon TV_{perm} \into Fl(n)$ induces a map
backwards on cohomology, which is neither injective nor surjective.
Klyachko \cite{klyachko1985orbits} presented its image
$im(\iota^*)$ (with rational coefficients), and a formula for $\iota^*$
evaluated on Schubert symbols:
\begin{eqnarray*}
  H^*(Fl(n);\QQ) &\to& im(\iota^*) \iso \QQ[k_0,\ldots,k_n]\bigg/
                   \left\langle
                   \begin{array}{c}
                     k_i(-k_{i-1}+2k_i-k_{i+1}) = 0,\ \  1\leq i \leq n-1 \\
                   k_0=k_n=0 
                   \end{array}\right\rangle \\                     
  \calS_\pi &\mapsto& \frac{1}{\ell(\pi)!} \sum_{Q \in RW(\pi)} \ \prod_{q\in Q} k_q
\quad\text{where $RW(\pi)$ is the set of reduced words}
\end{eqnarray*}

Taking forward- and back-stable limits, while leaving behind geometry,
we get the \defn{Klyachko genus}
$
  H(A_\ZZ) \to \QQ[\ldots,k_{-1},k_0,k_1,\ldots]\bigg/
                   \left\langle \
                     k_i(-k_{i-1}+2k_i-k_{i+1}) = 0\ \forall i\in \ZZ \
                   \right\rangle % \\                    
%  \calS_\pi &\to& \frac{1}{\ell(\pi)!} \sum_{Q:\ \prod Q = \pi} \ \prod_{q\in Q} k_q
$
whose map on Schubert symbols is given by the same formula.
%We consider this map as an analogue of $H(A_\ZZ) \to Symm$; each gives
%us a picture of back-stable Schubert calculus but inside another ring.
We use this to recover a result of Nenashev,
foreshadowing some results in \S\ref{sec:egg}:

\begin{thm}\cite[Proposition 3 and discussion after]{Nenashev}
\label{thm:Nenashev}
  Let $RW(\pi)$ denote the set of reduced words for $\pi$. There must
  {\bf exist} (but the proof doesn't find one) a ``rectification'' map
  $$ \{\text{shuffles of any word in $RW(\pi)$ with any word in $RW(\rho)$}\}
  \to \coprod_\sigma RW(\sigma) 
$$ \hskip -.05in % why the hell is this necessary here?
  whose fiber over any reduced word for $\sigma$ has
  size %cardinality
  $c_{\pi\rho}^\sigma$, the coefficient from
$ %  \begin{equation}    \label{eq:1}
    \calS_\pi \calS_\rho = \sum_\sigma c_{\pi\rho}^\sigma \, \calS_\sigma.
$ %  \end{equation}
\end{thm}

\begin{proof}
  Apply the Klyachko genus to that last equation, % (\ref{eq:1}),
  then set all $k_i=1$, obtaining
  $$
  \frac{1}{\ell(\pi)!}\sum_{P\in RW(\pi)}\prod_P 1 \ \ \ \
  \frac{1}{\ell(\rho)!}\sum_{R\in RW(\rho)}\prod_R 1
  =
  \sum_\sigma c_{\pi\rho}^\sigma \,
  \frac{1}{\ell(\sigma)!}\sum_{S\in RW(\sigma)}\prod_S 1
  $$
  Since $c_{\pi\rho}^\sigma = 0$ unless $\ell(\sigma) = \ell(\pi)+\ell(\rho)$,
  we can restrict to those $\sigma$. Multiplying through:
  $$
  \#RW(\pi)\ \#RW(\rho) {\ell(\pi)+\ell(\rho)\choose \ell(\pi)}
  =
  \sum_\sigma c_{\pi\rho}^\sigma \, \#RW(\sigma)
  $$ \hskip -.05in % why the hell is this necessary here?
  Let $C_{\pi\rho}^\sigma$ be a set with cardinality $c_{\pi\rho}^\sigma$
  (and wouldn't you like to know one?). Then we can interpret the above as
  $$ \#
   \{\text{shuffles of any word in $RW(\pi)$ with any word in $RW(\rho)$}\}
   \ =\ \# \coprod_\sigma (C_{\pi\rho}^\sigma \times RW(\sigma)) 
$$ \hskip -.05in % why the hell is this necessary here?
   Hence there exists a bijection; compose it with the
   projection to $\coprod_\sigma RW(\sigma)$.
\end{proof}

We can further simplify the target of this genus by modding
out by each of the minimal prime ideals that contain the Klyachko
ideal. We get ahold of these using the Nullstellensatz,\footnote{This
  isn't quite fair, in that we are working in infinite dimensions, but
  we won't worry about it. All we're really trying to do here is choose,
  for each $i$, which factor of $k_i(-k_{i-1}+2k_i-k_{i+1})$
  to mod out.}  i.e. by looking at the components of the
solution set to Klyachko's equations.

\begin{prop}\label{prop:Nullstellensatz}
  Consider $\ZZ$-ary tuples $(k_i)_{i\in \ZZ}$ of complex numbers satisfying
  the Klyachko equalities. This set is the (nondisjoint) union of the
  following countable set of $2$-planes:
  \begin{itemize}
  \item For $a,b \in \CC$,  let $k_m = am+b$.  \vskip-.5in \phantom{.}
  \item For $i\leq j$ each in $\ZZ$, and $x,y \in \CC$ a pair of
    ``slopes'', let
    $    k_m =    \begin{cases}
      x(m-i) & \text{if }k \leq i \\
      0 & \text{if }k \in [i,j] \\
      y(m-j) & \text{if }k \geq j.
    \end{cases}
    $
  \end{itemize}
\end{prop}

After completing this work, we learned of a very similar calculation
in \cite[\S 3.4]{NadeauTewari}, so we omit the proof of proposition
\ref{prop:Nullstellensatz} (obtainable as a sort of $q\to 1$ limit of theirs).

\junk{
\begin{proof}
  The equations say that each
  $k_m$ is either $0$ or is the average of its neighbors.
  Hence on any range $[i,j]$ over which $k_m\neq 0$, it is an
  affine-linear function of $m$. Also, if $k_j=0$ and $k_{j+1}=y\neq 0$,
  then by induction $k_{j+n} = ny \neq 0$ for $n\in\NN$, with
  a similar statement for $k_i=0$, $k_{i-1} = -x\neq 0$.
  Consequently, the set of $m$ s.t. $k_m=0$ forms an interval.

  Let $i = \inf(m: k_m=0)$ (or $+\infty$ if $\nexists m$),
  and $j = \sup(m: k_m=0)$ (or $-\infty$ if $\nexists m$).
  There are five cases:
  \begin{enumerate}
  \item $i = +\infty, j=-\infty$, i.e. $k_m$ is never $0$.
    Then $k_\bullet$ is an affine-linear function.
  \item $i \leq j$ are both finite. Then $k_\bullet$ is as described
    in the second case of the claim, where $x = -k_{i-1}$ and $y = k_{j+1}$.
  \item $i = -\infty$ but $j$ is finite. This is the $x=0$ case of \#2.
  \item $j = +\infty$ but $i$ is finite. This is the $y=0$ case of \#2.
  \item $i = -\infty$ and $j=+\infty$. This is the very trivial $x=y=0$ case
    of \#2.    \qedhere
  \end{enumerate}
\end{proof}
}

Each component defines a quotient of the Klyachko ring, namely
$$
\QQ[\ldots,k_{-1},k_0,k_1,\ldots]\bigg/
\left\langle
  -k_{m-1}+2k_m-k_{m+1} = 0\ \forall m\in \ZZ
\right\rangle
$$
$$
\forall i\leq j, \qquad
\QQ[\ldots,k_{-1},k_0,k_1,\ldots]\bigg/
\left\langle
  \begin{array}{cc}
    k_m = 0 &\forall m\in [i,j] \\
  -k_{m-1}+2k_m-k_{m+1} = 0& \forall m\notin [i,j]
  \end{array}
\right\rangle
$$ \hskip -.05in % why the hell is this necessary here?
Call the map of $H(A_\ZZ)$ to the first quotient the \defn{affine-linear genus}.

There is a slight subtlety in that the Klyachko ideal is not radical,
and as such, the map from the Klyachko ring to the direct sum of these
quotients is not injective. We will return to this minor matter below.

\subsection{Dropping the other genera}

The other components (besides the one giving the affine-linear genus)
are useless, in the following senses.  Say $k_m = 0$ for some $m$;
then there are three situations.
\begin{enumerate}
\item {\em Some reduced word for a permutation $\pi$ uses the letter  $m$.}
  Then {\em all} reduced words do, with the effect that $\calS_\pi \mapsto 0$
  in the quotient ring.
\item {\em Each reduced word for $\pi$ uses some letters $>m$ and some $<m$.}
  Then $\pi = \pi_{<m} \pi_{>m}$ where each
  uses only letters $<m$, $>m$ respectively. In this case
  $\calS_\pi = \calS_{\pi_{<m}} \calS_{\pi_{>m}}$.
\item {\em Each reduced word for $\pi$ only uses letters on one side of $m$.}
  At this point there is nothing to be gained by setting $k_m=0$; we could
  work with just the affine-linear genus.
\end{enumerate}
Our principal interest in genera is to study \defn{Schubert calculus},
the structure constants $c_{\pi\rho}^\sigma$
of the multiplication of Schubert symbols.
That is hard to do if the symbols map to zero (situation \#1), silly to do
directly if the symbols are are themselves products (situation \#2),
and in situation \#3 might as well be done using the affine-linear genus.
As such, at this point we cast aside the Klyachko genus in favor
of the affine-linear genus $\gamma$:
\begin{eqnarray*}
  \gamma:\ H(A_\ZZ) &\to& \QQ[a,b], \qquad \qquad\qquad
  \calS_\pi \mapsto \frac{1}{\ell(\pi)!}\ \sum_{P \in RW(\pi)}\ \prod_{i\in P} (ai+b)
\end{eqnarray*}

The assiduous reader might be guessing now that the information lost
when passing from the Klyachko ideal to its radical is similarly
negligible for Schubert calculus purposes. And indeed: if we factor the
Klyachko ideal as an intersection of primary instead of prime components,
we run into the ideals
$$
\forall i\leq j, \qquad
\QQ[\ldots,k_{-1},k_0,k_1,\ldots]\bigg/
\left\langle
  \begin{array}{cc}
    k_m^2 = 0 &\forall m\in [i+1,j-1] \\
    k_i = k_j = 0 &\\
  -k_{m-1}+2k_m-k_{m+1} = 0& \forall m\notin [i,j]
  \end{array}
\right\rangle
$$
\hskip -.05in % why the hell is this necessary here?
These would let us study $\pi,\rho,\sigma$ whose reduced words use only
the letters in the range $[i+1,j-1]$, and each at most once. This is an
extremely limited case.

\section{The affine-linear genus $\gamma$
  from the martial derivations}

Recall the derivations
$$ \nabla = \sum_m m \martial_m \qquad\qquad \xi = \sum_m \martial_m $$
\hskip -.05in % why the hell is this necessary here?
{\em Being} derivations, they exponentiate to automorphisms of
$\QQ \tensor_\ZZ H(A_\ZZ)$ (where the $\QQ$ is necessitated by
the denominators in the exponential series).

\begin{thm}\label{thm:triangle}
  The following triangle commutes: \quad
  $
  \begin{matrix}
    && H(A_\ZZ) \\ 
    & \qquad\qquad e^{a\nabla+b\xi} \swarrow  && \searrow \gamma \quad \quad \\ 
    & \QQ[a,b] \tensor_\ZZ H(A_\ZZ) & \onto & \QQ[a,b] \\ 
    & \calS_{\pi} & \mapsto & \delta_{\pi,e}
  \end{matrix}
  $
\end{thm}

\begin{proof}
  The proof is not conceptual; we compute both sides and compare.
  Indeed, we find the statement intriguing exactly because we know of
  no geometric reason the two maps should be related.
  $$ e^{a\nabla+b\xi} \cdot \calS_\pi
  \ =\ \sum_n \frac{1}{n!} (a\nabla+b\xi)^n \cdot \calS_\pi
  \ \mapsto\ \frac{ (a\nabla+b\xi)^{\ell(\pi)} \cdot \calS_\pi }{\ell(\pi)!}
  \ =\ \frac{ \left(\sum_i (ai+b)\martial_i\right) ^{\ell(\pi)} \cdot \calS_\pi }{\ell(\pi)!}
%  = \sum_n \frac{1}{n!} \sum_{S\subseteq [n]}
%  \prod_{i=1}^n (a\nabla)^{[i\in S]}(b\xi)^{[i\notin S]} \cdot \calS_\pi
  $$
  \hskip -.05in % why the hell is this necessary here?
  Expanding $\left(\sum_i (ai+b)\martial_i\right) ^{\ell(\pi)}$,
  the nonvanishing terms correspond to reduced words of length $\ell(\pi)$,
  and only those that multiply to $\pi^{-1}$ survive application to $\calS_\pi$.
\end{proof}

In particular the proof of Theorem \ref{thm:Nenashev} essentially
amounts to applying $\exp(\xi)$. (Oddly, the original proof in \cite{Nenashev}
is closer to an application of $\exp(\nabla)$.)

There is a fascinating \defn{$q$-Klyachko genus} introduced in
\cite[\S3.4]{NadeauTewari}:
\begin{eqnarray*}
  \gamma_q:\ H(A_\ZZ) &\to& \QQ(q)[\alpha,\beta] \\
  \calS_\pi &\mapsto& \frac{1}{\ell(\pi)\qtorial} \sum_{Q:\ \prod Q = \pi} \
                      q^{\comaj(Q)}  \prod_{i\in Q} \left(\alpha q^i+\beta\right)
\end{eqnarray*}
Here % $[i]_q = (q^i-1)/(q-1)$ is the $q$-number and
% (for $i>0$; if $i<0$ then $[i]_q = -q^{i+1}-q^{i+2}-\ldots-q^{-1}$)
$m\qtorial$ is the $q$-torial $\prod_{j=1}^m [j]_q$, and $\comaj(Q)$ is the sum of
the positions of the ascents. We looked for a long time for a $q$-analogue of
Theorem \ref{thm:triangle}, to no avail: it would provide an
automorphism of $H(A_\ZZ)(q)[\alpha,\beta]$
whose $\ell=0$ part is the $q$-Klyachko genus. 

\section{Rectification and the $q$-statistic}\label{sec:egg}

\junk{
We introduce a couple of extra variables into the $q$-Klyachko genus.
First, $H(A_\ZZ)$ is special among $H(D)$s for having a ``shift'' automorphism,
conjugating any $\pi$ by the permutation $i\mapsto i+1\ \forall i\in \ZZ$.
If we shift by $k$, we get
\begin{eqnarray*}
%   H(A_\ZZ) &\to& \QQ(q) \\
   H(A_\ZZ) \to \QQ(q), \qquad \qquad 
  \calS_\pi &\mapsto& \frac{1}{\ell(\pi)\qtorial} \sum_{Q:\ \prod Q = \pi} \
                      q^{\comaj(Q)}  \prod_{i\in Q} \frac{q^{i+k}-1}{q-1}
\end{eqnarray*}

Second, observe that any $H(D)$, being a $\ZZ$-graded ring, has an automorphism
multiplying each $\calS_\pi$ by $c^{\ell(\pi)}$. With this, we can
clear out the denominator from $\frac{q^{i+k}-1}{q-1}$, and the
$q^k$ in the numerator, but also bring in a factor $a$.
\begin{eqnarray*}
%   H(A_\ZZ) &\to& \QQ(q)[a] \\
   H(A_\ZZ) \to \QQ(q)[a], \qquad\qquad
  \calS_\pi &\mapsto& \frac{1}{\ell(\pi)\qtorial} \sum_{Q:\ \prod Q = \pi} \
                      q^{\comaj(Q)}  \prod_{i\in Q} (aq^i-aq^{-k})
\end{eqnarray*}
Now imagine replacing the $-aq^{-k}$ by a variable $b$. At each of
the (infinitely many) specializations $b := -aq^{-k}$, these rational
functions satisfy Monk's rule. Hence they do so for {\em all} $b$, and we can
take it as an extra variable.
\begin{eqnarray*}
%   H(A_\ZZ) &\to& \QQ(q)[a,b] \\
   H(A_\ZZ) \to \QQ(q)[a,b], \qquad\qquad
  \calS_\pi &\mapsto& \frac{1}{\ell(\pi)\qtorial}
                      \sum_{Q:\ \prod Q = \pi} \ q^{\comaj(Q)}  \prod_{i\in Q} (aq^i+b)
\end{eqnarray*}
} % end junk

We pursue a $q$-analogue of (Nenashev's) Theorem \ref{thm:Nenashev}.
Applying Nadeau-Tewari's $q$-Klyachko genus to 
$\calS_\pi \calS_\rho = \sum_\sigma c_{\pi\rho}^\sigma \, \calS_\sigma$
we get
\begin{eqnarray*}
&&  \frac{1}{\ell(\pi)\qtorial}
  \sum_{P\in RW(\pi)} \ q^{\comaj(P)}  \prod_{i\in P} (\alpha q^i + \beta) \quad
\frac{1}{\ell(\rho)\qtorial}
  \sum_{R\in RW(\rho)} \ q^{\comaj(Q)}  \prod_{i\in R} (\alpha q^i + \beta) \\
  &=&
\sum_\sigma c_{\pi\rho}^\sigma \, 
  \frac{1}{\ell(\sigma)\qtorial}
  \sum_{S\in RW(\sigma)} \ q^{\comaj(S)}  \prod_{i\in S} (\alpha q^i + \beta)
\end{eqnarray*}
Multiplying through, we get
\begin{eqnarray*}
  {\ell(\pi)+\ell(\rho)\choose \ell(\pi)}_q\
  \sum_{P \in RW(\pi)\atop R\in RW(\rho)} q^{\comaj(P)\atop +\comaj(R)} \prod_{i\in P\coprod R}(\alpha q^i + \beta)
  &=&
\sum_\sigma c_{\pi\rho}^\sigma \, 
  \sum_{S\in RW(\sigma)} \ q^{\comaj(S)}  \prod_{i\in S} (\alpha q^i + \beta)
\end{eqnarray*}

Let's interpret both sides at $\alpha = \beta = q = 1$, again using
a mystery set $C_{\pi\rho}^\sigma$ with cardinality $c_{\pi\rho}^\sigma$.
Define a \defn{barred word} for $\pi$ as a reduced word in which some
letters are overlined, e.g.  $12\overline 1$ for $(13)$.
Then the left side of the above equation counts pairs $(P,R)$ of
barred words, shuffled together, where the barring indicates ``use the
$\alpha q^i$ term'' rather than the $\beta$ term.  Meanwhile, the right side
counts pairs $(\tau,S)$ where $S$ is a barred word for some $\sigma$,
and $\tau$ is in $C_{\pi\rho}^\sigma$.

\begin{thm}\label{thm:qNenashev}
  Define the $q$-statistic of a barred word as the sum of the locations
  of the ascents, plus the sum of the barred letters.
  
  Define the $q$-statistic of a shuffle $\sha$ of a pair $(P,R)$
  of barred words as the sum of the two $q$-statistics, plus the number
  of inversions in the shuffle (letters in $R$ leftward of letters in $P$).

  Then there {\bf exists} (but the proof doesn't find one)
  a ``rectification'' map
  $$ \{\text{shuffles of pairs $(P,R)$ of barred words for $\pi,\rho$}\}
  \quad \to \quad \coprod_\sigma\ \{\text{barred words for $\sigma$}\}
  $$
  preserving the number of bars and the $q$-statistic, whose fiber
  over each word for $\sigma$ is of size $c_{\pi\rho}^\sigma$.
\end{thm}

We note that the affine-linear genus doesn't let one produce such a
combinatorial result, insofar as the factors $ai+b$ can involve $i<0$
(in the back-stable setting of $A_\ZZ$). 

{\em Example.} These examples get large very quickly, so we restrict to
the fully barred case. Let $\pi=\rho=12463578$, chosen to give a
$c_{\pi\rho}^\sigma>1$ (and chosen stably enough that the terms in the
product don't move $-\NN$). Each of $\pi$ and $\rho$ have two reduced
words ($354$ and $534$, $\comaj$s $1$ and $2$, each of total $12$),
and there are $6\choose 3$ ways to shuffle, for a total of
$2\cdot 2\cdot {6\choose 3} = 80$; the resulting $q$-statistics range
from $26 = 1+12+1+12+0$ to $37 = 2+12+2+12+3\cdot 3$.
There are $7$ terms $\calS_\sigma$ in the product
$\calS_\pi\calS_\rho$ (one with coefficient $2$) with various numbers
of reduced words.  \setcounter{MaxMatrixCols}{40}
$$
\begin{matrix}
  \text{$q$-statistic:} & 26 & 27 & 28 & 29 & 30 & 31 & 32 & 33 & 34 & 35 & 36 & 37\\ \hline
  & 1 & 3 & 5 & 8 & 11 & 12 & 12 & 11 & 8 & 5 & 3 & 1 &\text{total $=80$}\\
  \sigma\\
23561478 & 1 & 1 & 2 & 1 & 2 & 1 & 1 &   &   &   &   &  \\
14562378 &   & 1 &   & 1 & 1 & 1 &   & 1 &   &   &   &  \\
13572468 &   &   & 1 & 2 & 2 & 3 & 3 & 2 & 2 & 1 &   &  \\
13572468 &   &   & 1 & 2 & 2 & 3 & 3 & 2 & 2 & 1 &   &   &\text{again}\\
23471568 &   & 1 & 1 & 2 & 2 & 2 & 1 & 1 &   &   &   &  \\
13482567 &   &   &   &   & 1 & 1 & 2 & 2 & 2 & 1 & 1 &  \\
12673458 &   &   &   &   & 1 &   & 1 & 1 & 1 &   & 1 &  \\
12583467 &   &   &   &   &   & 1 & 1 & 2 & 1 & 2 & 1 & 1 
\end{matrix}
$$ \hskip -.05in % why the hell is this necessary here?
In the line with ``total $= 80$'', we count the number of fully barred
shuffles with given $q$-statistic. In each of the lower lines,
we put $\sigma$ on the left, and on the right
we list the number of fully barred words for it with given $q$-statistic.
Then the theorem asserts that each number atop a column is the sum of
the numbers below. There is a silly rotational near-symmetry tracing to the
fact that $\pi$ and $\rho$ are Grassmannian permutations for
self-conjugate partitions.

\section{Equidistribution of inversion number
  vs. $\comaj$ on $[n]\choose m$}

%\subsection{Before Monday Nov 6}


%{\em Example.}
Let $J \subseteq \ZZ$ be a set of $n$ numbers,
no two adjacent. Then the product $\prod_{j\in J} r_j$ is well-defined i.e$.$
is independent of the order; indeed, the reduced words for $\prod_{j\in J} r_j$
are in correspondence with permutations of $J$.
The same holds when multiplying subsets of $J$.

Fix $K\subseteq J$ and let $\rho = \prod_K r_k$,
$\pi = \prod_{J\setminus K} r_j$. 
Then $\calS_\pi \calS_\rho = \calS_{\pi \rho}$, and Theorem \ref{thm:qNenashev}
(again in the fully barred case) predicts a bijection
$$ \{\text{insertions of reduced words $R$ for $\rho$
  into reduced words $P$ for $\pi$} \}  \quad
\to \quad RW(\pi \rho)
$$
\hskip -.05in % why the hell is this necessary here?
such that $\big[\comaj(P)$ + $\comaj(R)$ + the inversion number of the
shuffle$\big]$
matches $\comaj$ of the resulting word $\sha$. Note that the obvious map (just
insert $R$ where the shuffle suggests) does {\em not} correspond
these two statistics!

If we break $J$ not into two subsets, but all the way down into individual
letters, this recovers the equidistribution on $S_n$ of the statistics
$\ell$ and $\comaj$ (or $\maj$); see e.g. \cite[Proposition 1.4.6]{EC1}.

This hints at a stronger result: that for any two strings $P,R$ such that $PR$ has no repeats,
on the set $\{$shuffles $\sha\}$ the distributions of the statistic
$\comaj(P) + \comaj(R) + \ell(\sha)$ and the statistic $\comaj(\sha)$ match.
(Theorem \ref{thm:qNenashev} only guarantees this {\em after summing}
over all $P\in RW(\pi),R\in RW(\rho)$.) %, so this conjecture is more precise.)
And indeed, this stronger claim holds \cite{GarsiaGessel}.

\junk{
junk all this later crap

More specifically, {\bf we give now a statistic-preserving map
that we conjecture to be bijective.} 

\junk{
Let $P,R$ be strings of numbers, each nonrepeating and with disjoint content.
In particular the shuffles of $R$ into $P$ are all distinct.
On the set $[\#P+\#R] \choose \#P$, which we interpret as partitions
$\lambda \subseteq \#R \times \#P$ inside a fixed rectangle,
we have the area statistic. We want to define a bijection
$$
\begin{matrix}
  \veccomaj :\
  &\{\text{shuffles of $R$ into $P$}\} \hskip -.4in \phantom{.}
  &\phantom{com}\to & {[\#P+\#R]\choose\#P} \phantom{area+\comaj(P)+\comaj(R)}\\
  &&\phantom{com}\searrow &\downarrow\text{area}+\comaj(P)+\comaj(R) \\
  &&\comaj\ \ \phantom.& \NN   \phantom{area+\comaj(P)+\comaj(R)}
\end{matrix}
$$ \hskip -.05in % why the hell is this necessary here?
making this triangle commute.

Our proposed rule is as follows. }
Fix $P$, $R$, and the shuffle $\sha$.
For $k\in [0,\#R]$ break $R$ as $R = R_{\leq k} R_{>k}$ where
$R_{\leq k}$ has the first $k$ letters of $R$. Let $P_k$ be
$P$ with $R_{>k}$ shuffled into it, i.e. as $k$ decreases from
$\#R$ to $0$ we shuffle $R$'s letters into $P$
one by one {\bf from the right} of $R$. For each $k$ we measure
$$ m_k := \comaj(P_k) - \comaj(R_{>k}), \quad k \in [0,\#R] 
$$ \hskip -.05in % why the hell is this necessary here?
so $m_{\#R} = \comaj(P)$ %\comaj(P_0) - \comaj(P) - \comaj(\emptyset) = 0
and $m_0 = \comaj(P_0) - \comaj(R)$. Then our rule is
to measure the jumps in $m_\bullet$, and sort them to get a partition
of $\comaj(P_0) - \comaj(R) - \comaj(P)$:
$$ \sha \quad\mapsto\quad {\mathrm sort}\left(m_{\#R-1} - m_{\#R},\ldots,
  m_1-m_2,  m_0-m_1\right) $$

{\em Example.} Let $P = 24$, $R = 315$ with $\comaj$s of $1,2$. The
(unsorted) vectors of jumps associated with the $5\choose 3$ shuffles are
$$
\renewcommand*{\arraystretch}{0}
\begin{matrix}
{}_{3}2_{15}4 \,\mapsto (0,0,0)   \\
& {}2_{315}4 \,\mapsto (0,0,1)   \\ 
{}24_{315} \,\mapsto (2,0,0)  & & {}_{31}2_{5}4 \,\mapsto (0,1,1)     \\
& {}_{3}24_{15} \,\mapsto (2,0,1)  & & {}_{315}24 \,\mapsto (1,1,1)     \\
{}2_{3}4_{15} \,\mapsto (2,0,2)  & & {}_{3}2_{1}4_{5} \,\mapsto (2,1,1)   \\
& {}2_{31}4_{5} \,\mapsto (2,1,2)     \\
{}_{31}24_{5} \,\mapsto (2,2,2) 
\end{matrix}
$$

\junk{
$P=4$, $R=361572$
$$
\begin{matrix}
{}_{36157}4_{2} \mapsto (0,0,0,0,0,0)   \\ 
{}_{3615}4_{72} \mapsto (0,1,0,0,0,0)    \\
{}_{36}4_{1572} \mapsto (0,1,1,0,0,0)   \\
4_{361572}      \mapsto (0,1,1,0,1,0)  \\
{}_{3}4_{61572} \mapsto (0,1,1,0,1,1)   \\
{}_{361}4_{572} \mapsto (0,1,1,1,1,1)  \\
{}_{361572}4   \mapsto (1,1,1,1,1,1)   
\end{matrix}
$$
}

We can prove our conjecture in the case $\#R=1$, where no sorting is
necessary. Attach $?$ and $\infty$ to $P$ making $?P\infty$, and for
each interstice in $?P\infty$ ask whether $P_m r P_{m+1}$ is an even
permutation (i.e. some cyclic rotation $abc$ has $a<b<c$) or odd;
here $?rP_1$ is considered to be both even and odd.
Then inserting $r$ at each odd
interstice from right to left followed by each even interstice left to right
produces shuffles whose $\comaj$s are $[\comaj\ P, \comaj\ P + \#P]$
in order. Here is an example, in which one follows the arrows to see
the possible $\comaj$s appearing in order.

$$
\begin{array}{cccccccc}
  \begin{matrix}     \comaj(P) + \\ \text{position of }5 \end{matrix}
  & \sha & \comaj(\sha)
  &\phantom.\phantom. &\quad
                             \begin{matrix}
                             \text{neighborhood of the }5 \end{matrix}
                             \quad\phantom.
&&\text{its parity} \\ 
\phantom 1 8 & {\bf 5}1 3 2 7 8 6 4 &  11 && \phantom 3 51 & 
        \ \ursh&\text{``both"}&\rdsh\  \\
\phantom 1 9 &  1{\bf 5}3 2 7 8 6 4 &  10 && 153 & \uparrow&\text{odd} \\
10 &   1 3{\bf 5}2 7 8 6 4 &  12 && 352 & &\text{even}&\downarrow \\
11 &   1 3 2{\bf 5}7 8 6 4 &  13 && 257 & &\text{even}&\downarrow \\
12 &   1 3 2 7{\bf 5}8 6 4 &  \phantom 1 9 && 758 &\uparrow&\text{odd} \\
13 &   1 3 2 7 8{\bf 5}6 4 &  14 && 856  & &\text{even}&\downarrow \\
14 &   1 3 2 7 8 6{\bf 5}4 &  \phantom 1 8 && 654 & \uparrow&\text{odd} \\
15 &   1 3 2 7 8 6 4{\bf 5}&  15 && 45\mathrlap\infty\phantom 8  & &\text{even}&\downarrow 
\end{array}
$$




\junk{
\subsubsection{Older}

We construct $\lambda := \veccomaj(\sha)$ one row at a time, with a twist:
sometimes we determine $\lambda_1$, other times $\lambda_{\#R}$,
and then continue inductively.

The dichotomy is as follows. Let $r$ be the rightmost value in $R$.
Pad $P$ in front and back as $?P\infty$, and label each interstice
in $?P\infty$ that $r$ might shuffle into with
\begin{itemize}
\item \defn{even} if $r$ and its adjacent neighbors $p,p'$ have
  $prp'$ in cyclic order (like $123$, $231$, or $312$), or $p=?$
  (giving the trio the benefit of the doubt), and
\item \defn{odd} if $r$ and its adjacent neighbors $p,p'$ have
  $prp'$ cyclically backwards (like $213$, $132$, or $321$), or $p=?$
  (again giving the trio the benefit of the doubt). In particular
  the first interstice (left of $P$, right of the $?$) gets both labels.
\end{itemize}
Finally the rule:
\begin{itemize}
\item if $r$ is shuffled into an even interstice, let
  $\lambda_1 = \#P$ minus the number of even interstices to the right.
  With that $1\times \lambda_1$ row removed we have a smaller
  partition to determine inductively.
\item if $r$ is shuffled into an odd interstice,
  let $\lambda_{\#R} = $ the number of odd interstices to the right.
  With that $\#R \times \lambda_{\#R}$ {\em rectangle} removed we have a
  smaller partition to determine inductively.
\end{itemize}
Either way, cut $P$ at the interstice, attach the last element of $R$ there
(instead of $\infty$), and continue inductively as described above. 
}
} % end junk from midsection

\acknowledgements{We thank Marcelo Aguiar, Hugh Dennin, Yibo Gao, Thomas Lam,
  Philippe Nadeau, Gleb Nenashev, Mario Sanchez, Thomas B\r a\r ath
  Sj\"oblom, and David E Speyer.}

\junk{
\section{Proof attempt}


\begin{lem}
  Let $\#P=n$, $\#R=m$. Let $\sha$ be a shuffle of $R$ into $P$.
  \begin{enumerate}
  \item If $R_m$ is placed immediately left of $P_n$ in the shuffle,
    then $\lambda(P,R)$ is $\lambda(P_1\cdots P_{n-1} R_m, R_1\cdots R_{m-1})$
    blah blah.
  \item If $R_m$ is placed further left than that,
    then $\lambda(P,R) = \lambda(P_1\cdots P_{n-1}, R)$
    if $P_{n-1}>P_n$, or is $\lambda(P_1\cdots P_{n-1}, R)$
    plus an entire column (height $m$) attached if $P_{n-1}<P_n$.
  \end{enumerate}
\end{lem}

\begin{proof}
  
\end{proof}

\begin{thm}
  Consider the shuffles of $R$ into $P$, but whose last letter is
  that of $P$ not $R$. Let $n = \#P$, $m = \#R$. 
  To each such shuffle we associate a partition,
  as in the conjecture. There are two cases:
  \begin{enumerate}
  \item[i.] If $R_m < P_n$, we obtain once each partition inside
    $\#R \times \#P$ that avoids the last column.
  \item[ii.] If $R_m > P_n$, we obtain once each partition inside
    $\#R \times \#P$ that contains the first column.
  \end{enumerate}
\end{thm}

\begin{proof}
  Let $\Sha_{P,R}$ denote the set of shuffles considered.
  Then there is a bijection
  $$ \beta\colon\quad
  \Sha_{P,R} \ \to\ \Sha_{P_1\cdots P_{n-1}R_m,\ R_1\cdots R_{m-1}}  \sqcup\
  \Sha_{P_1\cdots P_{n-1},\ R}   
  $$
  \hskip -.05in % why the hell is this necessary here?
  where $\sha \in \Sha_{P,R}$ either takes $R_m$ immediately to the
  left of $P_n$ (in which case we replace $P_n$ with $R_m$,
  and continue to insert the rest of $R$), vs. further to the left (in
  which case we can safely remove $P_n$). This bijection doesn't quite
  commute with the computation $\sha\mapsto \lambda(\sha)$ of the partition;
  where it does not, we write ``$1+$'' to indicate adding a column on
  the left.  There are several cases (indicated by {\em i} and {\em ii}
  as in the theorem statement).
  $$
  \begin{array}{c|ccc}
    & \lambda(\Sha_{P,R}) & 
    \lambda( \Sha_{P_1\cdots P_{n-1}R_m,\ R_1\cdots R_{m-1}} ) &
    \lambda( \Sha_{P_1\cdots P_{n-1},\ R}) \\ \hline
  \end{array}
  $$
\end{proof}

} %end junk

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