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\keywords{Quantum group, Fock space, symmetric groups, double covers, RoCK blocks}
\subjclass{17B37, 05E10, 20C25, 20C30}
\begin{document}
\title[Comparing Fock spaces in types $A^{(1)}$ and $A^{(2)}$]{Comparing Fock spaces\\in types $A^{(1)}$ and~$A^{(2)}$}
\author{\firstname{Matthew} \lastname{Fayers}}
\address{Queen Mary University of London\\Mile End Road\\London E1 4NS\\U.K.}
\email{m.fayers@qmul.ac.uk}
\thanks{This research was partly supported by EPSRC Small Grant EP/W005751/1.}
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\begin{abstract}
We compare the canonical bases of level-$1$ quantised Fock spaces in affine types $A^{(1)}$ and $A^{(2)}$, showing how to derive the canonical basis in type $A^{(2)}_{2n}$ from the the canonical basis in type $A^{(1)}_n$ in certain weight spaces. In particular, we derive an explicit formula for the canonical basis in extremal weight spaces, which correspond to RoCK blocks of double covers of symmetric groups. In a forthcoming paper with Kleshchev and Morotti we will use this formula to find the decomposition numbers for RoCK blocks of double covers with abelian defect.
\end{abstract}
\maketitle
\section{Introduction}
This paper is motivated by the \emph{decomposition number problem} for the symmetric groups and their double covers in characteristic $p$. Although a solution to this problem seems a long way off, several important results are known. One of these results gives the decomposition numbers for \emph{RoCK blocks} of symmetric groups; these are particularly well understood blocks which have been used in a variety of applications. The formula for the decomposition numbers for RoCK blocks in the abelian defect case was given by Chuang and Tan \cite{cts}, and results of Turner \cite{turn} allow these results to be extended to RoCK blocks with non-abelian defect groups.
It is natural to seek analogous results for the double cover $\hsss n$ of the symmetric group (which controls projective representations of $\sss n$). The case of characteristic $2$ behaves very differently (and is dealt with in \cite{mfspin2alt}), so we concentrate here on odd characteristic. The representations of $\hsss n$ which do not descend to representations of $\sss n$ are called \emph{spin} representations of $\sss n$, and the blocks of $\hsss n$ containing spin representations are called spin blocks. RoCK blocks for symmetric groups can be characterised as elements of the maximal equivalence class of blocks under the Scopes equivalence \cite{sco} on blocks of symmetric groups. The Scopes--Kessar equivalence \cite{kes} for spin blocks of double covers suggests a natural analogue of RoCK blocks, and this can be realised in a combinatorial way using the abacus. These blocks have recently been studied in detail by Kleshchev and Livesey \cite{kl}, who prove Brou\'e's abelian defect group conjecture for RoCK blocks. This has been used even more recently by Brundan and Kleshchev, and independently by Ebert, Lauda and Vera, to show that Brou\'e's conjecture holds for all spin blocks \cite{bk,elv}. However, the results of Kleshchev and Livesey do not directly address the decomposition number problem for spin RoCK blocks, and this is the main focus here.
In this paper we address RoCK blocks by studying quantum algebra. Let $U=U_q(A^{(1)}_{p-1})$ denote the quantised universal enveloping algebra of the affine Kac--Moody algebra of type $A^{(1)}_{p-1}$. The \emph{level-$1$ Fock space} is a highest-weight $U$-module with a simple combinatorial construction in terms of integer partitions. The submodule generated by a highest-weight vector is isomorphic to the irreducible highest-weight module $V(\La_0)$, so the Fock space provides a combinatorial framework for studying $V(\La_0)$; this approach has proved useful, for example, in constructing crystals. $V(\La_0)$ possess an important basis called the \emph{canonical basis}, which provides a connection to representation theory of symmetric groups and Iwahori--Hecke algebras, via the work of Lascoux--Leclerc--Thibon \cite{llt} and Ariki \cite{ari}, who showed that decomposition numbers for Hecke algebras of type $A$ in characteristic zero can be obtained by specialising canonical basis coefficients at $q=1$. This means in particular that these decomposition numbers can be calculated algorithmically. A further conjecture due to James suggested that the same should apply for decomposition numbers of symmetric groups, in blocks with abelian defect groups. James's conjecture is now known to be false in general \cite{willi}, but there are a wide variety of situations where it is known to hold, and it has provided inspiration for results on decomposition numbers. In particular, the formula due to Chuang and Tan \cite{cts} for decomposition numbers for RoCK blocks of $\sss n$ was inspired by their earlier calculation of the canonical basis in weight spaces corresponding to RoCK blocks, and shows in particular that James's conjecture holds for RoCK blocks.
An analogous connection to quantum groups for spin representations was found by Leclerc and Thibon \cite{lt}, using the quantum group of type $A^{(2)}_{p-1}$. This quantum group also acts on a combinatorially defined level-$1$ Fock space (now defined in terms of $p$-strict partitions), which possesses an irreducible highest-weight submodule with a canonical basis. Leclerc and Thibon formulated an analogue of James's conjecture for spin representations of $\hsss n$. This conjecture is also known not to hold in general, but it does hold in many special cases, in particular all known cases of decomposition numbers for RoCK blocks. Motivated by this conjecture, the aim of the present paper is to find the canonical bases for the weight spaces of the basic $U_q(A^{(2)}_{p-1})$-module corresponding to spin RoCK blocks. Rather than directly determining the canonical basis, we deduce our result from the results of Chuang--Tan by proving more general results comparing the canonical bases in types $A^{(1)}$ and $A^{(2)}$: we show that if $\be$ is a restricted $p$-strict partition satisfying a particular additional condition which says that the $p$-bar-core of $\be$ is large in a certain specific sense relative to the sum of the parts of $\be$ divisible by $p$, then we can obtain the canonical basis element labelled by $\be$ from a corresponding canonical basis element in type $A^{(1)}$ by an adjustment involving inverse Kostka polynomials. In certain cases this allows the canonical bases for entire weight spaces to be computed, including weight spaces corresponding to RoCK blocks. Combining this result with the Chuang--Tan formula for RoCK blocks in type $A^{(1)}$ yields our main result (\cref{mainrouq}).
Combining our theorem with the Leclerc--Thibon conjecture (specialised to the case of RoCK blocks), we arrive at a conjecture for the decomposition numbers for RoCK blocks (\cref{rockconj}). In a forthcoming paper with Kleshchev and Morotti \cite{fkm} we prove this conjecture.
\section{Background}
In this section we set out some background details on partitions and Fock spaces.
\subsection{Elementary notation}
We write $\bbn$ for the set of positive integers and $\bbn_0=\bbn\cup\{0\}$. Given $m\in\bbn$, the set $\zmz$ consists of cosets $a+m\bbz=\lset{a+mb}{b\in\bbz}$. Given any set $B\subseteq\bbz$ and $a\in\bbz$, we write $B+a=\lset{b+a}{b\in B}$.
\subsection{Partitions}
A \emph{partition} is an infinite weakly decreasing sequence $\la=(\la_1,\la_2,\dots)$ of non-negative integers which is eventually zero. When writing partitions, we omit the trailing zeroes and group together equal parts with a superscript. The partition $(0,0,\dots)$ is written as $\vn$. If $\la$ is a partition, the integers $\la_1,\la_2,\dots$ are called the \emph{parts} of $\la$. We write $\card\la=\la_1+\la_2+\cdots$, and we say that $\la$ is a partition of $\card\la$. The \emph{length} $\len\la$ is the number of non-zero parts of $\la$. We write $\calp$ for the set of all partitions.
If $\la$ is a partition and $n\in\bbn$, then $n\la$ is defined to be the partition $(n\la_1,n\la_2,\dots)$. If $\la$ and $\mu$ are partitions, then $\la\sqcup\mu$ is defined to be the partition of $\card\la+\card\mu$ obtained by combining all the parts of $\la$ and $\mu$ in decreasing order.
The \emph{Young diagram} of $\la$ is the set
\[
\lset{(r,c)\in\bbn^2}{c\ls\la_r}
\]
whose elements are called the \emph{nodes} of $\la$. In general, a node means an element of $\bbn^2$. We use the English convention for drawing Young diagrams, in which $r$ increases down the page and $c$ increases from left to right. We abuse notation by identifying $\la$ with its Young diagram; so for example we may write $\la\subseteq\mu$ to mean that $\la_r\ls\mu_r$ for all $r$.
If $\la$ is a partition, the \emph{conjugate} partition $\la'$ is the partition obtained by reflecting the Young diagram of $\la$ in the main diagonal; that is, $\la'_r=\card{\lset{c\in\bbn}{\la_c\gs r}}$.
The \emph{dominance order} on partitions is defined by writing $\la\dom\mu$ (and saying that $\la$ \emph{dominates} $\mu$) if $\card\la=\card\mu$ and $\la_1+\dots+\la_r\gs\mu_1+\dots+\mu_r$ for all $r$.
We say that a node $(r,c)$ of $\la$ is \emph{removable} if it can be removed from~$\la$ to leave the Young diagram of a partition (that is, if $c=\la_r>\la_{r+1}$), and we write the resulting partition as $\la\setminus(r,c)$. Similarly, a node $(r,c)$ not in $\la$ is an \emph{addable node} of $\la$ if it can be added to $\la$ to give a partition, and we write this partition as $\la\cup(r,c)$.
Now fix an integer $m\gs2$. We say that a partition $\la$ is \emph{$m$-restricted} if $\la_r-\la_{r+1}r$, the number of bar-removable nodes in row $s$ equals the number of bar-addable nodes in row $s+1$, except when $\al_s\equiv1\ppmod h$, in which case there is one more bar-removable node in row $s$. Furthermore, there is an $n$-bar-addable node in row $s$ \iff $\al_s\equiv n\ppmod h$. So if we let $a_1$ be the number of $s>r$ such that $\al_s\equiv1\ppmod h$, and define $a_n$ similarly, then
\[
\ip{\spin g\al}{\be}=q^{2(a_n-a_1+1)}.
\]
(The extra $1$ arises from the bar-addable node in row $r+1$.)
But now observe that $\al_s\equiv n\ppmod h$ \iff $(\phi\al)_s-s\in\hi-1$, while $\al_s\equiv1\ppmod h$ \iff $(\phi\al)_s-s\in\hi$. So $a_n-a_1+1$ equals the number of $i$-addable nodes of $\phi\la$ below row $r$ minus the number of $\hi$-removable nodes (the extra $1$ arises from the addable node in row $l+1$). So $\ip{\spin g\la}{\be}=\ip{\f\hi\phi\al}{\phi\be}$.
\end{proof}
This gives an analogue of \cref{simpleg}. We continue to write $\hi=-l+n\bbz$.
\begin{cory}\label{hardq}
Suppose $\al$ is a \nice \bp, and that $\f\hi\phi\al\in\fo^{\ls l}$. Then $\spin g\al\in\fonicel$, and
\[
\Phi(\spin g\al)=\f\hi\phi\al.
\]
\end{cory}
\begin{proof}
From the fact that (1)$\Rightarrow$(2) in \cref{hardg} we can write
\[
\spin g\al=\sum_{\be\in\pnice l}c_\be\be
\]
for some coefficients $c_\be\in\bbc(q)$, so that $\spin g\al\in\fonicel$. Then the last statement in \cref{hardg} tells us that for each $\mu\in\pl l$ the coefficient of $\mu$ in $\f\hi\phi\al$ is $c_{\phi^{-1}\mu}$. By assumption no $\mu$ with $\mu\notin\pl l$ occurs in $\f\hi\phi\al$, and so
\[
\f i\phi\al=\sum_{\mu\in\pl l}c_{\phi^{-1}\mu}\mu=\sum_{\be\in\pnice l}c_\be\Phi(\be)=\Phi(\spin g\al).\qedhere
\]
\end{proof}
Now we can prove our first main result.
\begin{proof}[Proof of \cref{samedec}]
Take a restricted partition $\be\in\pnice l$, and let $\mu=\phi\be$. The LLT algorithm \cite[Section~6.2]{llt} shows that $\cb n(\mu)$ can be written as a linear combination $\sum_\nu a_\nu A(\nu)$, where:
\begin{enumerate}
\item\label{nudommu}
the sum is over a set of partitions $\nu$ with $\nu\dom\mu$;
\item
each coefficient $a_\nu$ lies in $\bbc(q+q^{-1})$;
\item
each $A(\nu)$ has the form $\f{i_1}\dots \f{i_r}\vn$ for some $i_1,\dots,i_r\in\znz$;
\item\label{xidomnu}
each $A(\nu)$ is a linear combination of partitions $\xi$ with $\xi\dom\nu$.
\end{enumerate}
In particular, conditions (\ref{nudommu}) and (\ref{xidomnu}) imply that each $A(\nu)$ lies in $\fo^{\ls l}$. So if we take such a $\nu$ and write
\[
A(\nu)=\f{i_1}\dots \f{i_r}\vn=\sum_{\xi\in\pl l}c_{\nu\xi}\xi
\]
with each $c_{\nu\xi}\in\bbc(q)$, then by \cref{simpleg,hardq} we can find $j_1,\dots,j_s\in\{0,\dots,n\}$ and a coefficient $b_\nu\in\bbc(q+q^{-1})$ such that
\[
\Phi^{-1}(A(\nu))=b_\nu\spf{j_1}\dots\spf{j_s}\phi^{-1}\vn=\sum_{\xi\in\pl l}c_{\nu\xi}\phi^{-1}\xi.
\]
(The coefficient $b_\nu$ arises because of the divided power occurring in the definition of the operator $\spin g$.) Hence the vector
\[
\Phi^{-1}(\cb n(\mu))=\sum_\nu a_\nu\Phi^{-1}(A(\nu))
\]
can be written as a linear combination of vectors of the form $\spf{j_1}\dots\spf{j_s}\phi^{-1}\vn$, with coefficients in $\bbc(q+q^{-1})$. Now $\phi^{-1}\vn$ is an $h$-bar-core: its parts are simply the smallest $l$ positive integers whose residues modulo $h$ lie in $\{1,\dots,n\}$. So $\phi^{-1}\vn=\scb h(\phi^{-1}\vn)$ is bar-invariant, and so $\Phi^{-1}(\cb n(\mu))$ is bar-invariant. Since $\Phi^{-1}(\cb n(\mu))=\sum_\la \dn\la\mu\phi^{-1}\la$, with $\dn\mu\mu=1$ and all other $\dn\la\mu$ divisible by $q$, the uniqueness of the canonical basis means that $\Phi^{-1}(\cb n(\mu))=\scb h(\be)$.
\end{proof}
\section{Comparing types $A^{(1)}_{n-1}$ and $A^{(1)}_{m-1}$: runner addition}\label{runaddsec}
In this section we invoke a theorem from the author's paper \cite{mfrunrem} comparing canonical bases in types $A^{(1)}_{n-1}$ and $A^{(1)}_{m-1}$, and use it to express \cref{samedec} in terms of the canonical basis for $\calv_m$; this will also allow us to include the case $h=3$.
Let $\pml$ denote the set of partitions $\la$ with length at most $l$ such that $\la_r+l+1-r\nnequiv0\ppmod l$ for $1\ls r\ls l$. Another way of saying this is that $\la\in\pml$ if $\la$ can be displayed on an $m$-runner abacus with $l+1$ beads such that runner $0$ contains only a bead in position $0$. We define a bijection $\pl l\to\pml$ which we denote $\la\mapsto\la^+$, by setting
\[
\la^+_r=\la_r+\inp{\frac{\la_r+l-r}n}
\]
for $r=1,\dots,l$. This bijection is easily visualised on the abacus: given $\la\in\pl l$, we take the $n$-runner \abd for $\la$ with $l$ beads, add a runner at the left with a bead in the top position only, and let $\la^+$ be the resulting partition. It is easy to see that $\la^+$ is $m$-restricted \iff $\la$ is $n$-restricted. Extending the map $\la\mapsto\la^+$ linearly, we obtain an injective linear map $\fo^{\ls l}\to\fo$ written $v\mapsto v^+$.
Now we can give the main result from \cite{mfrunrem}. Some translation of notation is required, since in \cite{mfrunrem} the opposite convention for $\fo$ is used, in which the canonical basis elements $\cb n(\mu)$ are indexed by $n$-regular partitions, and a ``full'' runner is added rather than an empty one. But replacing all partitions with their conjugates yields the following result.
\begin{thm}[\xcite{mfrunrem}{Theorem 3.1}]\label{runadd}
Suppose $n\gs2$ and $\la\in\pl l$ is $n$-restricted. Then $\cb m(\la^+)=\cb n(\la)^+$.
\end{thm}
We combine this with the results of the last section to give a result comparing canonical bases in types $A^{(1)}_{m-1}$ and $A^{(2)}_{h-1}$. Recalling the bijection $\phi:\pnice l\to\pl l$ from Section~\ref{firstcomparesec}, let $\psi:\pnice l\to\pml$ be the bijection defined by $\psi\la=(\phi\la)^+$. Explicitly, $\psi$ is given by
\[
(\psi\al)_r=\al_r-(m-1)\inp{\frac{\al_r}h}-l+r-1
\]for $1\ls r\ls l$.
We define an injective $\bbc(q)$-linear map $\Psi:\fonicel\to\fo$ by $\Psi(\be)=\psi\be$. Now we obtain the following result.
\begin{thm}\label{firstmain}
Suppose $\be\in\pnice l$ is restricted. Then
\[
\cb m(\psi\be)=\Psi(\scb h(\be)).
\]
\end{thm}
\begin{proof}
If $h\gs5$ then this follows immediately from Theorems \ref{samedec} and \ref{runadd}. In the case $h=3$, the assumption that $\be$ is restricted means that $\be$ is the partition $(3l-2,3l-5,\dots,1)$, which is a $3$-bar-core, so that $\scb3(\be)=\be$. On the other hand, $\psi\be$ is the partition $(l,l-1,\dots,1)$ which is a $2$-core, so that $\cb2(\psi\be)=\psi\be$. So the result follows in this case too.
\end{proof}
\section{Comparing types $A^{(1)}_{m-1}$ and $A^{(2)}_{h-1}$: separated partitions}\label{nextbitsec}
Having established \cref{firstmain}, we will work with $U_m$ rather than $U_n$ from now on. With this in mind, if $a$ is an integer then we may write $\f a$ to mean $\f{a+m\bbz}$.
Our next aim is to extend \cref{firstmain} to include $h$-strict partitions with positive parts divisible by $h$. To do this, we need to restrict attention to what we call \emph{separated} partitions. In this section we define separated partitions and prove two preparatory results giving similar calculations in $\fo$ and~$\spin\fo$.
\subsection{Separated partitions}\label{sepsec}
We keep $l\in\bbn$ fixed. Given a partition $\la\in\pml$ and any partition $\pi$, we define a partition $\jo\la\pi$ as follows: we take the \abd for $\la$ with $m$ runners and $am+l+1$ beads for sufficiently large $a$, and then move the $r$th lowest bead on runner $0$ down $\pi_r$ positions for each $r$. Another way to express this is as follows: we write $\la=\mu^+$ for $\mu\in\pl l$; then we join together the $1$-runner \abd for $\pi$ with $a+1$ beads and $n$-runner \abd for $\mu$ with $an+l$ beads. It is easy to see that $\jo\la\pi$ is independent of the choice of $a$.
Given such an \abd for $\jo\la\pi$, let $b$ be the position of the last bead on runner $0$ (that is, $b=(\pi_1+a)m$), and let $f$ be the first empty position not on runner $0$. Now given $k\in\bbn_0$, we say that $\jo\la\pi$ is \emph{$k$-\sep} if $f-b>km$. For $k=0$, we will just say \emph{\sep} rather than $0$-\sep. Whether $\jo\la\pi$ is $k$-\sep is independent of the choice of $a$ (in fact, it only depends on $l$, $k$, $\pi_1$ and $\la'_1$).
\begin{example}
Take $h=7$, so that $m=4$, and let $l=10$. If we take $\la=(7,6^2,4,1^3)$ and $\pi=(1^2)$, then $\jo\la\pi=(7,6^2,4,1^{11})$, as we see from the following \abds.
\[
\begin{array}{c@{\qquad}c}
\abacus(lmmr,bbbb,bbbb,bbbb,nbbb,nnnb,nnbb,nbnn)
&
\abacus(lmmr,bbbb,nbbb,bbbb,bbbb,nnnb,nnbb,nbnn)
\\[38pt]
\la
&
\jo\la\pi
\end{array}
\]
We obtain $b=12$ and $f=17$, so that $\jo\la\pi$ is $1$-\sep but not $2$-\sep.
\end{example}
Now for each $k\in\bbn$ we define an operator $F_k$ on the Fock space by
\[
F_k=\f{-l}^{(k)}\f{1-l}^{(k)}\dots \f{m-1-l}^{(k)}.
\]
The action of the operator $F_k$ can be visualised using the abacus: given an \abd with $l+am+1$ beads for a partition $\la$, we apply $F_k$ by moving $k$ beads from runner $m-1$ to runner $0$, then $k$ beads from runner $m-2$ to runner $m-1$, and so on, and finally moving $k$ beads from runner $0$ to runner $1$, where at each stage a bead moves from position $b$ to position $b+1$ for some $b$. The partitions that can be obtained in this way are the partitions that appear in $F_k\la$.
The first result we want to prove describes $F_k(\jo\la\pi)$ when $\jo\la\pi$ is $k$-\sep. Recall that given two partitions $\pi,\rho$, we write $\pi\car\rho$ if $\rho$ can be obtained from $\pi$ by adding $r$ nodes in distinct columns.
\begin{propn}\label{addrun1simple}
Suppose $\la\in\pml$, $\pi\in\calp$ and $k,u\in\bbn_0$, and that $\jo\la\pi$ is $(k+u)$-\sep. If $\nu\in\calp$, then $\ip{F_k(\jo\la\pi)}{\nu}\neq0$ \iff $\nu$ has the form $\jo\mu\rho$, where
\begin{itemize}
\item
$\mu\in\pml$ and $\rho\in\calp$,
\item
$\pi\car\rho$ for some $0\ls r\ls k$, and
\item
$\ip{F_{k-r}\la}{\mu}\neq0$.
\end{itemize}
Furthermore, if these conditions hold, then $\jo\mu\rho$ is $u$-\sep.
\end{propn}
\begin{proof}
In this proof we use an abacus with $m$ runners and $l+am+1$ beads for fixed large $a$. For any partition $\nu$ we write $\ab\nu$ for the $m$-runner \abd for $\nu$ with $l+am+1$ beads.
Given a finite set $B$ of positions on the abacus, we use the phrase \emph{making bead moves from $B$} to mean moving a bead from position $b$ to position $b+1$ for each $b\in B$.
Suppose $\nu$ appears in $F_k(\jo\la\pi)$ with non-zero coefficient. From the description of the action of the $\f i$ on $\fo_m$ (interpreted in terms of the abacus) there are sets $B_0,\dots,B_{m-1}\subset\bbn_0$ with $\card{B_g}=k$ and $B_g\subset g+m\bbz$ for each $g$, such that $\ab\nu$ is obtained from $\ab{\jo\la\pi}$ by making bead moves from $B_{m-1},B_{m-2},\dots,B_0$ in turn. We write $B=B_0\cup\dots\cup B_{m-1}$.
Let $xm$ be the position of the last bead on runner $0$ in $\ab{\jo\la\pi}$; that is, $x=\pi_1+a$. The assumption that $\jo\la\pi$ is $(k+u)$-\sep means that every position before position $(x+k+u)m$ not on runner $0$ is occupied in $\ab{\jo\la\pi}$. We make two observations about the relationships between the sets $B_0,\dots,B_{m-1}$.
\begin{enumerate}[label=(\roman*)]
\item
If $b\in B_0$ with $b>xm$, then $b-1\in B_{m-1}$ (otherwise it is not possible to make a bead move from $b$ at the final step).
\item\label{underx}
If $b\in B_g$ with $0\ls gym
\end{cases}
\end{align*}
which is injective, and hence bijective. Now we partition each $B_g$ as $C_g\cup D_g$, where
\[
C_g=\lset{b\in B_g}{bym}.
\]
The fact that $f$ is bijective together with observation \ref{underx} above means that
\[
C_g=C_0+g\qquad\text{ for }0\ls g\ls m-1,\tag{\tac}\label{lsy}
\]
so that in particular $\card{C_0}=\dots=\card{C_{m-1}}=r$, say. The fact that $f$ is bijective also gives
\[
D_{m-1}+1=D_0.\tag{\tac\tac}\label{gsy}
\]
So the combined bead moves from $D_{m-1},\dots,D_0$ have no effect on runner $0$. So (\ref{lsy}) implies that runner $0$ of $\ab\nu$ is obtained from runner $0$ of $\ab{\jo\la\pi}$ by adding a bead at position $b+m$ for all $b\in C_0$, and then removing a bead from position $b$ for all $b\in C_0$. In particular, there are exactly $a$ beads on runner $0$ in $\ab\nu$, and the last bead on runner $0$ is in position $ym$ or earlier. (\ref{lsy}) also means that all the positions before position $ym$ on runners other than runner $0$ are occupied in $\ab\nu$. This means that $\nu$ has the form $\jo\mu\rho$ with $\mu\in\pml$ and $\rho\in\calp$. Now we compare the beta-sets $\ber{a+1}\pi$ and $\ber{a+1}\rho$. If we let $A=\lset{c/m+1}{c\in C_0}$, then the above description of runner $0$ of $\ab\nu$ means that
\[
\card A=r,\qquad A\cap\ber{a+1}\pi=\emptyset,\qquad \ber{a+1}\rho=\ber{a+1}\pi\cup A\setminus(A-1).
\]
So $\pi\car\rho$ by Lemma~\ref{carbeta}(2).
Another consequence of (\ref{lsy}) is that the combined bead moves from $C_{m-1},\dots,C_0$ have no effect on any runner other than runner $0$. This means that if we take $\ab\la$ and make the bead moves from $D_{m-1},\dots,D_0$ in turn, then we obtain $\ab\mu$. So $\mu$ appears in $F_{k-r}\la$ with non-zero coefficient.
Now we show that $\jo\mu\rho$ is $u$-\sep. Let $D=D_0\cup\dots\cup D_{m-1}$. Because $\pi\car\rho$, the last bead on runner $0$ in $\ab{\jo\mu\rho}$ is in position $(x+r)m$ or earlier. Let $d$ be the first empty position in $\ab{\jo\mu\rho}$ not on runner $0$. If $d>(x+k+u)m$, then certainly $\jo\mu\rho$ is $u$-\sep, so assume $d<(x+k+u)m$. Then position $d$ is occupied in $\ab{\jo\la\pi}$, so $d\in D$. Now observation \ref{underx} and (\ref{gsy}) imply that $d+1,d+2,\dots,(x+k+u)m$ all belong to $D$ as well. Hence
\[
m(k-r)=\card D>(x+k+u)m-d,
\]
giving $d>(x+r+u)m$, so that $\jo\mu\rho$ is $u$-\sep.
\smallskip
Now we need to prove the ``if'' part of the \lcnamecref{addrun1simple}. So suppose we are given $\mu,\rho,r$ satisfying the conditions in the \lcnamecref{addrun1simple}. Because $\mu$ appears in $F_{k-r}\la$ with non-zero coefficient, $\ab\mu$ is obtained from the \abd for $\la$ by making bead moves from $D_{m-1},\dots,D_0$ in turn, where $D_g\subset g+m\bbz$ and $\card{D_g}=k-r$ for each $g$. In addition, $D_0=D_{m-1}+1$ because $\la$ and $\mu$ have identical configurations on the $0$th runner. In the same way as in the last paragraph, we can show that every element of $D_0\cup\dots\cup D_{m-1}$ is greater than $(x+r+u)m$.
The assumption that $\pi\car\rho$ means (using Lemma~\ref{carbeta}(2)) that $\ber{a+1}\rho=\ber{a+1}\pi\cup A\setminus(A-1)$ for some set $A$ with $\card A=r$ and $A\cap\ber{a+1}\pi=\emptyset$. So if we define
\[
C_g=\lset{(a-1)m+g}{a\in A}
\]
for each $g$, then $\ab{\jo\mu\rho}$ is obtained from $\ab{\jo\la\pi}$ by making bead moves from $C_{m-1}\cup D_{m-1},\dots,C_0\cup D_0$ in turn, so $\jo\mu\rho$ appears in $F_k(\jo\la\pi)$.
\end{proof}
Now we want to show that the coefficients appearing in \cref{addrun1simple} coincide.
\begin{propn}\label{addrun1}
Suppose $\la,\mu\in\pml$ and $\pi,\rho\in\calp$ with $\pi\car\rho$ for some $r$, and that $k\in\bbn$ is such that $k\gs r$ and $\jo\la\pi$ is $k$-\sep. Then
\[
\ip{F_k(\jo\la\pi)}{\jo\mu\rho}=\ip{F_{k-r}\la}{\mu}
\]
\end{propn}
\begin{proof}
\cref{addrun1simple} shows that one side of the given equation is non-zero \iff the other is, so we assume they are both non-zero. We use an abacus with $l+am+1$ beads, and we define the integer $y$ and the sets $B_g=C_g\cup D_g$ for $0\ls g\ls m-1$ as in the proof of \cref{addrun1simple}. Then $\mu$ is obtained from $\la$ by making bead moves from $D_{m-1},\dots,D_0$ in turn, and $\jo\mu\rho$ is obtained from $\jo\la\pi$ by making bead moves from $B_{m-1},\dots,B_0$ in turn. We remark that these sets are uniquely defined: if $\jo\mu\rho$ appears in $F_k(\jo\la\pi)=\f{-l}^{(k)}\f{1-l}^{(k)}\dots \f{-1-l}^{(k)}(\jo\la\pi)$, then we obtain $\jo\mu\rho$ by adding all the $(-1-l)$-nodes of $(\jo\mu\rho)\setminus(\jo\la\pi)$, then all the $(-2-l)$-nodes of $(\jo\mu\rho)\setminus(\jo\la\pi)$, and so on, with no choice at any stage.
So if we define partitions $\lb\la0,\dots,\lb\la m$ by setting $\lb\la0=\la$ and defining $\lb\la g$ from $\lb\la{g-1}$ by making bead moves from $D_{m-g}$ for $g\gs1$, then $\lb\la m=\mu$ and
\[
\ip{F_{k-r}\la}{\mu}=\prod_{g=1}^m\ip{\f{-g-l}^{(k-r)}\lb\la{g-1}}{\lb\la g},
\]
and the formula for the action of $\f{-g-l}^{(k-r)}$ says that
\[
\ip{\f{-g-l}^{(k-r)}\lb\la{g-1}}{\lb\la g}=q^{2n(\lb\la{g-1},\lb\la g)}.
\]
Similarly we define partitions $\lb\xi0,\dots,\lb\xi m$ by setting $\lb\xi0=\jo\la\pi$ and then making bead moves from $B_{m-1},\dots,B_0$, and derive a similar formula for $\ip{F_k(\jo\la\pi)}{\jo\mu\rho}$. We need to compare $n(\lb\la{g-1},\lb\la g)$ with $n(\lb\xi{g-1},\lb\xi g)$ for each $g$.
Taking $2\ls g\ls m-1$ first and letting $i=-g-l+m\bbz$, we compare the addable and removable $i$-nodes of $\lb\la{g-1},\lb\la g,\lb\xi{g-1},\lb\xi g$, by looking at runners $m-g$ and $m-g+1$ of their \abds. For any given row we may give the configuration of beads on the two positions on runner $m-g$ and $m-g+1$ in that row.
The rows on these runners where these four partitions differ are the rows containing elements of $C_{m-g}$, where
\begin{itemize}
\item
$\lb\la{g-1}$ and $\lb\la g$ both have $\miabb$,
\item
$\lb\xi{g-1}$ has $\miabn$,
\item
$\lb\xi g$ has $\mianb$,
\end{itemize}
and the rows containing elements of $D_{m-g}$, where
\begin{itemize}
\item
$\lb\la{g-1}$ and $\lb\xi{g-1}$ both have $\miabn$,
\item
$\lb\la g$ and $\lb\xi g$ both have $\mianb$.
\end{itemize}
We conclude that $\lb\la{g-1}$ and $\lb\xi{g-1}$ have exactly the same removable $i$-nodes while $\lb\la g$ and $\lb\xi g$ have exactly the same addable $i$-nodes. Moreover, none of the addable $i$-nodes of $\lb\xi g$ or removable $i$-nodes of $\lb\xi{g-1}$ lie to the left of any of the added nodes corresponding to the positions in $C_{m-g}$. So $n(\lb\la{g-1},\lb\la g)=n(\lb\xi{g-1},\lb\xi g)$, which means that $\ip{\f i^{(k-r)}\lb\la{g-1}}{\lb\la g}=\ip{\f i^{(k)}\lb\xi{g-1}}{\lb\xi g}$.
Now we consider the case $g=1$, by comparing runners $m-1$ and $0$ of $\lb\la0,\lb\la1,\lb\xi0,\lb\xi1$. To help in visualising addable and removable nodes, we imagine runner $m-1$ moved to lie to the left of runner $0$, and shifted down so that position $cm-1$ is directly to the left of position $cm$ for each $c$. So for a given partition when we show the abacus configuration on these runners in row $c$, we show positions $cm-1$ and $cm$. (We temporarily introduce a new position $-1$, which is taken to be occupied in all \abds.)
After position $ym$, the partitions $\lb\la0$ and $\lb\xi0$ agree, as do the partitions $\lb\la1$ and $\lb\xi1$. Now we look at the rows before position $ym$.
The \abds for $\lb\la0$ and $\lb\la1$ both have $\miabb$ in the first $a+1$ rows, and $\miabn$ in the remaining $y-a$ rows up to position $ym$.
Now consider the positions $cm-1$ and $cm$ in the \abds for $\lb\xi0$ and $\lb\xi1$, where $cm\ls y$. The configurations in these positions are determined by the sets $C_{m-1}$ and $\ber{a+1}\pi$:
\begin{itemize}
\item
if $cm-1\in C_{m-1}$, then $\lb\xi0$ has $\miabn$ and $\lb\xi1$ has $\mianb$;
\item
if $cm-1\notin C_{m-1}$ and $c\notin\ber{a+1}\pi$, then $\lb\xi0$ and $\lb\xi1$ both have $\miabn$;
\item
if $c\in\ber{a+1}\pi$, then $\lb\xi0$ and $\lb\xi1$ both have $\miabb$.
\end{itemize}
Note in particular that the second possibility happens exactly $y-a-r$ times. This enables us to compare $n(\lb\la0,\lb\la1)$ with $n(\lb\xi0,\lb\xi1)$ by examining addable and removable nodes. We find that
\[
n(\lb\xi0,\lb\xi1)=n(\lb\la0,\lb\la1)-(k-r)r+|A|,
\]
where
\[
A=\lset{(c,d)}{0\ls ckh$. (We write simply ``\bsep'' to mean ``$0$-\bsep''.)
Now for each $k\in\bbn$, define $\spin F_k=\spf0^{(2k)}\spf 1^{(2k)}\dots\spf{n-1}^{(2k)}\spf n^{(k)}$.
Then we can give an analogue of \cref{addrun1simple}.
\begin{propn}\label{addrun2simple}
Suppose $\al\in\pnice l$, $\pi\in\calp$ and $k,u\in\bbn_0$, and that $\jh\al\pi$ is $(k+u)$-\bsep. If $\ga\in\hstr$, then $\ip{\spin F_k(\jh\al\pi)}{\ga}\neq0$ \iff $\ga$ has the form $\jh\be\rho$, where
\begin{itemize}
\item
$\be\in\pnice l$ and $\rho\in\calp$,
\item
$\pi\car\rho$ for some $0\ls r\ls k$, and
\item
$\ip{\spin F_{k-r}\al}{\be}\neq0$.
\end{itemize}
Furthermore, if these conditions hold, then $\jh\be\rho$ is $u$-\bsep.
\end{propn}
\begin{proof}
We follow the structure of the proof of \cref{addrun1simple}, though the details here are slightly different. For any $h$-strict partition $\ga$, we write $\bab\ga$ for the \babd of $\ga$. As in the proof of \cref{addrun1simple}, given a finite set $B\subset\bbn_0$, we use the phrase \emph{making bead moves from $B$} to mean moving a bead from position $b$ to position $b+1$ for each $b\in B$.
Suppose $\ga$ appears in $\spin F_k(\jh\al\pi)$. Let $\be$ be the partition obtained from $\ga$ by removing all the positive parts divisible by $h$, and let $\rho$ be the partition obtained by taking all the parts of $\ga$ divisible by $h$ and dividing them all by $h$. The assumption that $\ip{\spin F_k(\jh\al\pi)}{\ga}\neq0$ means that we can obtain $\bab\ga$ from $\bab{\jh\al\pi}$ by making bead moves from
\[
B_n,\ B_{n+1}\cup B_{n-1},\ B_{n+2}\cup B_{n-2},\ \dots,\ B_{h-1}\cup B_0
\]
in turn, where each $B_g$ is a set of non-negative integers congruent to $g$ modulo $h$, and $\card{B_n}=k$ while $\card{B_{n+g}}+\card{B_{n-g}}=2k$ for $1\ls g\ls n$.
Let $x=\pi_1$. Then the lowest bead on runner $0$ in $\bab{\jh\al\pi}$ is in position $xh$. Because $\card{B_n}=k$, we can choose $y\in\{x,\dots,x+k\}$ such that $yh+n\notin B_n$. This has two consequences.
\begin{itemize}
\item
Because $\jh\al\pi$ is $k$-\bsep, positions $yh+1,\dots,yh+n$ are occupied in $\bab{\jh\al\pi}$. The assumption that $yh+n\notin B_n$ then means that $yh+g\notin B_g$ for $g=1,\dots,n-1$.
\item
Because $\al$ is \nice, runners $n+1,\dots,h-1$ are empty in $\bab{\jh\al\pi}$ and position $(y+1)h$ is unoccupied. The assumption that $yh+n\notin B_n$ then means that $yh+g\notin B_g$ for $g=n+1,\dots,h-1$, and $(y+1)h\notin B_0$.
\end{itemize}
Now write $B_g=C_g\cup D_g$ for each $g$, where
\[
C_g=\lset{b\in B_g}{b\ls yh},\qquad D_g=\lset{b\in B_g}{b>yh}.
\]
The fact that runners $n+1,\dots,h-1$ are empty in $\bab{\jh\al\pi}$, and that there are no beads on runner $0$ after position $yh$, means that $C_{n+g}\subseteq C_{n+g-1}+1$ and $D_{n+g}\subseteq D_{n+g-1}+1$ for $g=1,\dots,n$, and also that $D_0\subseteq D_n+n+1$. In addition, the fact that all the positions on runners $1,\dots,n$ before position $yh$ are occupied in $\bab{\jh\al\pi}$ means that $C_{n-g}\subseteq C_{n-g+1}-1$ for $g=1,\dots,n$. So
\[
\card{C_0}\ls\dots\ls\card{C_n}\gs\dots\gs\card{C_{h-1}}
\]
and
\[
\card{D_n}\gs\dots\gs\card{D_{h-1}}\gs\card{D_0}
\]
Now the fact that $\card{C_{h-1}}+\card{C_0}+\card{D_{h-1}}+\card{D_0}=2k=2\card{C_n}+2\card{D_n}$ gives equality everywhere. So $C_g=C_0+g$ for $g=1,\dots,h-1$ and $D_{n+g}=D_n+g$ for $g=1,\dots,n$, and $D_0=D_n+n+1$.
So the combined bead moves from $C_n,C_{n+1}\cup C_{n-1},\dots,C_{h-1}\cup C_0$ only affect runner $0$ in the \babd, where the effect is to add a bead at position $b+h$ for each $b\in C_0$ and then remove a bead from position $b$ for each $b\in C_0$. By Lemma~\ref{carbeta}(1), this means that $\pi\car\rho$.
Now we look at the other runners in $\bab\ga$. From what we have learned about the sets $C_g$ and $D_g$, we know that there are $l$ beads on runners $1,\dots,n$, and no beads on runners $n+1,\dots,h-1$. So $\be\in\pnice l$, and $\ga=\jh\be\rho$. Furthermore, $\bab\be$ is obtained from $\bab\al$ by making bead moves from $D_n,D_{n+1}\cup D_{n-1},\dots,D_{h-1}\cup D_0$, and $\card{D_g}=k-r$ for each $g$, so $\ip{\spin F_{k-r}\al}{\be}\neq0$.
We also need to show that $\jh\be\rho$ is $u$-\bsep; this is done in the same way as the corresponding part of \cref{addrun1simple}.
\smallskip
Now we prove the ``if'' part of the \lcnamecref{addrun2simple}. Suppose we are given $\be,\rho,r$ as in the \lcnamecref{addrun2simple}. Because $\ip{\spin F_{k-r}\al}{\be}\neq0$, there are sets $D_0,\dots,D_{h-1}\subseteq\bbn_0$ with $D_g\subset g+h\bbz$ for each $g$ and $\card{D_{n+g}}+\card{D_{n-g}}=2(k-r)$ for $g=0,\dots,n$, such that $\bab\be$ is obtained from $\bab\al$ by making bead moves from $D_n,D_{n+1}\cup D_{n-1},\dots,D_{h-1}\cup D_0$ in turn. Because $\al$ and $\be$ are both \nice, we obtain $D_{n+g}=D_n+g$ for $g=1,\dots,n$, so that in fact $\card{D_g}=k-r$ for every $g$.
The assumption that $\pi\car\rho$ means that there is a set $A\subset\bbn$ with $\card A=r$ such that $\rho$ is obtained from $\pi$ by adding a part equal to $a$ for every $a\in A$, and then removing a part equal to $a-1$ for each $a\in A$. So if we define
\[
C_g=\lset{(a-1)h+g}{a\in A},\qquad B_g=C_g\cup D_g
\]
for each $g$, then $\bab{\jh\be\rho}$ is obtained from $\bab{\jh\al\pi}$ by making bead moves from $B_n,B_{n+1}\cup B_{n-1},B_{h-1}\cup B_0$ in turn, so $\jh\be\rho$ appears in $\spin F_k(\jh\al\pi)$.
\end{proof}
Now we compare the coefficients appearing in \cref{addrun2simple}. This is similar to \cref{addrun1}, though here the statement is more complicated. Given partitions $\pi,\rho$ with $\pi\car\rho$, we let $\hsb\pi\rho$ be defined as in Section~\ref{newpierisec}, \emph{with the indeterminate $t$ replaced by $-q^2$}; that is,
\[
\hsb\pi\rho=\prod_{\substack{c\gs1\\\rho'_c=\pi'_c\\\rho'_{c+1}>\pi'_{c+1}}}(1-(-q^2)^{\pi'_c-\pi'_{c+1}}).
\]
\begin{propn}\label{addrun2}
Suppose $\al,\be\in\pnice l$ and $\pi,\rho\in\calp$ with $\pi\car\rho$ for some $r$, and that $\jh\al\pi$ is $k$-\bsep for some $k\gs r$. Then
\[
\ip{\spin F_k(\jh\al\pi)}{\jh\be\rho}=\hsb\pi\rho\ip{\spin F_{k-r}\al}{\be}.
\]
\end{propn}
\begin{proof}
\cref{addrun2simple} shows that one side of the given equation is non-zero \iff the other is, so we assume they are both non-zero. We define the integer $y$ and the sets $B_g=C_g\cup D_g$ for $0\ls g\ls h-1$ as in the proof of \cref{addrun2simple}. Then $\be$ is obtained from $\al$ by making bead moves from $D_n,D_{n+1}\cup D_{n-1},\dots,D_{h-1}\cup D_0$ in turn, and $\jh\be\rho$ is obtained from $\jh\al\pi$ by making bead moves from $B_n,B_{n+1}\cup B_{n-1},\dots,B_{h-1}\cup B_0$ in turn. As in the proof of \cref{addrun1}, these sets are uniquely defined.
Define partitions $\lb\al0,\dots,\lb\al m$ by letting $\lb\al0=\al$, and then constructing $\lb\al{g+1}$ from $\lb\al g$ by making bead moves from $B_{n+g}\cup B_{n-g}$, for $g=0,\dots,n-1$. Then $\lb\al m=\be$, and
\begin{alignat*}2
\ip{\spin F_{k-r}\al}{\be}&=\ip{\spf n^{(k-r)}\al}{\lb\al1}\ \mathrlap{\times\ \prod_{g=1}^n\ip{\spf{n-g}^{(2(k-r))}\lb\al g}{\lb\al{g+1}}.}
\\
\intertext{Correspondingly,}
\ip{F_k(\jh\al\pi)}{\jh\be\rho}&=\ip{\spf n^{(k)}(\jh\al\pi)}{\jh{\lb\al1}\pi}\ &&\times\ \prod_{g=1}^{n-1}\ip{\spf{n-g}^{(2k)}(\jh{\lb\al g}\pi)}{\jh{\lb\al{g+1}}\pi}
\\
&&&\times\ \ip{\spf0^{(2k)}(\jh{\lb\al n}\pi)}{\jh\be\rho}.
\end{alignat*}
First we compare the terms
\[
\ip{\spf n^{(k-r)}\al}{\lb\al1}=q^{4\spin n(\al,\lb\al1)},
\qquad
\ip{\spf n^{(k)}(\jh\al\pi)}{\jh{\lb\al1}\pi}=q^{4\spin n(\jh\al\pi,\jh{\lb\al1}\pi)}.
\]
We can compute the integers $\spin n(\al,\lb\al1)$ and $\spin n(\jh\al\pi,\jh{\lb\al1}\pi)$ by looking at runners $n$ and $n+1$ in the bar-abacus. As in the proof of \cref{addrun1}, we may draw a small diagram showing the configuration of beads on these runners in a given row. These two runners are identical in the four partitions $\al,\lb\al1,\jh\al\pi,\jh{\lb\al1}\pi$, except:
\begin{itemize}
\item
in the positions corresponding to elements of $C_n$, where $\al$, $\lb\al1$ and $\jh\al\pi$ all have $\mibbn$ while $\jh{\lb\al1}\pi$ has $\mibnb$;
\item
in the positions corresponding to elements of $D_n$, where $\al$ and $\jh\al\pi$ have $\mibbn$ while $\lb\al1$ and $\jh{\lb\al1}\pi$ have $\mibnb$.
\end{itemize}
Now comparing the calculation of $\spin n(\al,\lb\al1)$ and $\spin n(\jh\al\pi,\jh{\lb\al1}\pi)$, we obtain
\[
\spin n(\jh\al\pi,\jh{\lb\al1}\pi)=\spin n(\al,\lb\al1)-(k-r)r+\card A,
\]
where
\[
A=\lset{0\ls c\pi'_{c+1}$ while $\rho'_c=\pi'_c$. As a result, $N_2=\hsb\pi\rho$, as required.
\end{proof}
\section{Comparing canonical bases}\label{comparingsec}
In this section we give our main results comparing canonical bases. In order to use the results of Sections~\ref{sepsec} and ~\ref{barsepsec}, we need to compare the actions of the operators $F_k$ and $\spin F_k$ defined in Section~\ref{nextbitsec}. We begin with the following \lcnamecref{samecoeff}.
\begin{propn}\label{samecoeff}
Suppose $\al,\be\in\pnice l$. Then
\[
\ip{\spin F_k\al}{\be}=\ip{F_k\psi\al}{\psi\be}.
\]
\end{propn}
\begin{proof}
For this proof we write $\la=\psi\al$ and $\mu=\psi\be$. First we show that one side of the equation is non-zero \iff the other is. We use the terminology and notation relating to the (bar)-abacus from the proofs of \cref{addrun1simple,addrun1,addrun2simple,addrun2}; we use \abds with $l+1$ beads for $\la$ and $\mu$. We define functions
\begin{alignat*}2
\hf:\bbz&\longrightarrow\bbz&\qquad\fh:\bbz&\longrightarrow\bbz
\\
b&\longmapsto b+(m-h)\inp{\frac bh}&b&\longmapsto b+(h-m)\inp{\frac bm}.
\end{alignat*}
Observe that $\hf$ and $\fh$ restrict to mutually inverse bijections between $g+h\bbz$ and $g+m\bbz$ for each $0\ls g\ls n$, and that position $b$ is occupied in $\ab\la$ \iff position $\fh(b)$ is occupied in $\bab\al$ (and similarly for $\mu$ and $\be$).
Suppose first that $\ip{\spin F_k\al}\be\neq0$. Then there are sets $D_0,\dots,D_{h-1}\subset\bbn_0$ with $D_g\subset g+h\bbz$ and $\card{D_{n-g}}+\card{D_{n+g}}=2k$ for each $g$, such that $\bab\be$ is obtained from $\bab\al$ by making bead moves from $D_n,D_{n+1}\cup D_{n-1},\dots,D_{h+1}\cup D_0$ in turn. The assumption that $\al,\be\in\pnice l$ implies that $D_{n+g}=D_{n+g-1}+1$ for $g=1,\dots,n$, and that $D_0=D_{h-1}+1$. So in fact $\card{D_g}=k$ for each $g$. Now if we define $E_g=\hf(D_g)$ for $g=0,\dots,n$, then $E_g\subset g+m\bbz$ and $\card{E_g}=k$ for each $g$, and $\ab\mu$ can be obtained from $\ab\la$ by making bead moves from $E_n,E_{n-1},\dots,E_0$ in turn. So $\ip{F_k\la}\mu\neq0$.
The converse is very similar: if $\ip{F_k\la}\mu\neq0$, then there are sets $E_0,\dots,E_n$ with the properties given above and satisfying $E_0=E_n+1$. Now if we define $D_g=\fh(E_g)$ for $g=0,\dots,n$, and $D_g=D_{g-1}+1$ for $g=n+1,\dots,h-1$, then the sets $D_g$ have the properties in the above paragraph and $\bab\be$ is obtained from $\bab\al$ by making bead moves from $D_n,D_{n+1}\cup D_{n-1},\dots,D_{h-1}\cup D_0$. So $\ip{\spin F_k\al}\be\neq0$.
So we assume that both sides of the equation are non-zero, and define the sets $D_0,\dots,D_{h-1}$ and $E_0,\dots,E_n$ as in the last two paragraphs. Also, for $g=1,\dots,n$ let $F_g$ be the set of parts $b$ of $\al$ congruent to $g$ modulo $h$ such that $b\notin D_g$; in other words, $F_g$ is the set of positions of beads on runner $g$ of $\bab\al$ that are \emph{not} moved in constructing $\be$ from $\al$. We define the partitions $\lb\al0,\dots,\lb\al m$ as in the proof of \cref{addrun2}, and the partitions $\lb\la0,\dots,\lb\la m$ as in the proof of \cref{addrun1} (using $E_0,\dots,E_n$ in place of $D_0,\dots,D_n$) , so that
\begin{align*}
\ip{\spin F_k\al}{\be}&=\ip{\spf n^{(k)}\al}{\lb\al1}\ \times\ \prod_{g=1}^n\ip{\spf{n-g}^{(2k)}\lb\al g}{\lb\al{g+1}},
\\
\ip{F_k\la}{\mu}&=\prod_{g=0}^n\ip{\f{-g-1-l}^{(k)}\lb\la g}{\lb\la{g+1}}.
\end{align*}
We will compare these two expressions term by term. First consider $\ip{\spf n^{(k)}\al}{\lb\al1}=q^{4\spin n(\al,\lb\al1)}$ and $\ip{\f{-1-l}^{(k)}\la}{\lb\la1}=q^{2n(\la,\lb\la1)}$. Since runner $n+1$ of $\bab\al$ is empty, $\al$ has no $n$-bar-removable nodes, so
\[
\spin n(\al,\lb\al1)=\card{\lset{(c,d)}{c0$ with $\ip{F_k(\jo\la\pi)}{\jo\mu\rho}\neq0$. If $\jo\la\pi$ is $(k+u)$-\ssep for some $u\gs0$, then $\jo\mu\rho$ is $u$-\ssep.
\end{lemma}
\begin{proof}
The statement that $\jo\la\pi$ is $(k+u)$-\ssep is the same as saying that $\jo\la\pi$ is $(k+u+\card\pi-\pi_1)$-\sep. Since $\jo\mu\rho$ appears with non-zero coefficient in $F_k(\jo\la\pi)$, \cref{addrun1simple} shows that $\jo\mu\si$ does as well, where $\si$ is the partition obtained by adding $\card\rho-\card\pi$ nodes to $\pi$ at the end of the first row. So from the final statement of \cref{addrun1simple}, $\jo\mu\si$ is $(u+\card\pi-\pi_1)$-\sep, which is the same as $(u+\card\si-\si_1)$-\sep, so $\jo\mu\si$ is $u$-\ssep, and therefore $\jo\mu\rho$ is $u$-\ssep.
\end{proof}
Now for each $u$ define
\begin{align*}
\sspu u&=\lspan{\jo\la\pi}{\la\in\pml,\ \pi\in\calp,\ \jo\la\pi\text{ $u$-\ssep}},
\\
\bsspu u&=\lspan{\jh\al\pi}{\al\in\pnice l,\ \pi\in\calp,\ \jh\al\pi\text{ $u$-\sbsep}}.
\end{align*}
We write $\ssp$ and $\bssp$ for $\sspu0$ and $\bsspu0$.
Lemma~\ref{fssep} shows that $F_k$ maps $\sspu{k+u}$ to $\sspu u$. Similarly, $\spin F_k$ maps $\bsspu{k+u}$ to $\bsspu u$. To extend \cref{samecoeff} to connect the actions of $F_k$ and $\spin F_k$, we need to define a linear bijection between $\bssp$ and $\ssp$. Recalling the Kostka polynomials $K_{\rho\pi}(t)$ from Section~\ref{symfnsec}, we define
\[
\asa:\bssp\longrightarrow\ssp
\]
by setting
\[
\asa(\jh\al\pi)=\sum_{\rho\in\calp}K_{\rho\pi}(-q^2)\jo{\psi\al}\rho
\]
and extending linearly. ($\asa$ is bijective because the matrix of Kostka polynomials $K_{\rho\pi}(t)$ for $\rho,\pi\in\calp(r)$ is invertible for each $r$.)
Now we can reconcile \cref{addrun1,addrun2} via the following \lcnamecref{sasfk}.
\begin{propn}\label{sasfk}
Suppose $v\in\bsspu k$. Then $\asa(\spin F_kv)=F_k\asa(v)$.
\end{propn}
\begin{proof}
By linearity we can assume $v=\jh\al\pi$ for some $k$-\sbsep partition $\jh\al\pi$. Since both sides of the equation lie in $\ssp$, it suffices to show that
\[
\ip{\asa(\spin F_k(\jh\al\pi))}{\jo\mu\si}=\ip{F_k\asa(\jh\al\pi)}{\jo\mu\si}
\]
for each \sep partition $\jo\mu\si$.
We use the results of Section~\ref{symfnsec}, with the variable $t$ specialised to $-q^2$. So we let $P_\la,Q_\la$ be the two types of Hall--Littlewood symmetric functions defined for $t=-q^2$, and $\lan\,,\,\ran$ the Hall--Littlewood inner product.
\allowdisplaybreaks
Now
\begin{align*}
\ip{\asa(\spin F_k(\jh\al\pi))}{\jo\mu\si}
&=\sum_{\substack{\tau\in\calp\\\pi\car\tau}}\sum_{\be\in\pnice l}\hsb\pi\tau\ip{\spin F_k\al}{\be}\ip{\asa\left(\jh\be\tau\right)}{\jo\mu\si}
\tag*{(by \cref{addrun2})}
\\
&=\sum_{\substack{\tau\in\calp\\\pi\car\tau}}K_{\si\tau}(-q^2)\hsb\pi\tau\ip{\spin F_k\al}{\psi^{-1}\mu}
\tag*{(from the definition of $\asa$)}
\\
&=\sum_{\substack{\tau\in\calp\\\pi\car\tau}}\lan s_\si,Q_\tau\ran\hsb\pi\tau\ip{F_k\psi\al}{\mu}
\tag*{(by \cref{samecoeff} and the definition of the coefficients $K_{\si\tau}(t)$)}
\\
&=\sum_{\tau\in\calp}\lan s_\si,Q_\tau\ran\lan\partial_{(r)}P_\tau,Q_\pi\ran\ip{F_k\psi\al}{\mu}
\tag*{(by \cref{dualpieri})}
\\
&=\lan\partial_{(r)}s_\si,Q_\pi\ran\ip{F_k\psi\al}{\mu}
\tag*{(since $s_\si=\sum_\tau\lan s_\si,Q_\tau\ran P_\tau$)}
\\
&=\sum_{\substack{\rho\in\calp\\\rho\car\si}}\lan s_\rho,Q_\pi\ran\ip{F_k\psi\al}{\mu}
\tag*{(by the Pieri rule for Schur functions)}
\\
&=\sum_{\rho\in\calp}K_{\rho\pi}(-q^2)\ip{F_k\jo{\psi\al}\rho}{\jo\mu\si}
\tag*{(by \cref{addrun1})}
\\
&=\ip{F_k\asa(\jh\al\pi)}{\jo\mu\si}
\tag*{(from the definition of $\asa$).\qedhere}
\end{align*}
\end{proof}
Now we want to use \cref{sasfk} to compare canonical basis elements. Given partitions $\pi,\rho$, we write $\rho\succcurlyeq\pi$ if either $\card\rho<\card\pi$ or $\rho\dom\pi$. Part (2) of the next \lcnamecref{sscbv} is our main result comparing canonical bases.
\begin{thm}\label{sscbv}
Suppose $\al\in\pnice l$ and $\pi\in\calp$, and that $\jh\al\pi$ is $u$-\sbsep for some $u\gs0$. Then:
\begin{enumerate}
\item
$\scb h(\jh\al\pi)$ is a linear combination of basis elements $\jh\be\rho$ in which $\rho\succcurlyeq\pi$ and $\jh\be\rho$ is $u$-\sbsep;
\item
$\asa(\scb h(\jh\al\pi))=\cb m(\jo{\psi\al}\pi)$.
\end{enumerate}
\end{thm}
\newcommand\vc V
\begin{proof}
We proceed by induction on $\pi$, using the order $\succcurlyeq$. The case $\pi=\vn$ follows from \cref{firstmain}, so assume $\pi\neq\vn$, and that the theorem is true if $\pi$ is replaced by any $\rho$ with $\rho\succ\pi$.
Let $k$ be the last non-zero part of $\pi$, let $\pi^-$ denote the partition obtained by deleting this last part, and consider the bar-invariant vector $\vc=\spin F_k\scb h(\jh\al{\pi^-})$. Since $\jh\al{\pi^-}$ is $(k+u)$-\sbsep, the inductive hypothesis says that $\scb h(\jh\al{\pi^-})$ is a linear combination of basis elements $\jh\ga\si$ for which $\si\succ\pi^-$ and $\jh\ga\si$ is $(k+u)$-\sbsep. Now \cref{addrun2simple} shows that the terms $\jh\be\rho$ appearing in $\vc$ all have the properties that $\si\car\rho$ for some $r\ls k$ and $\jh\be\rho$ is $u$-\sbsep. If $\si\car\rho$ with either $\card\si<\card{\pi^-}$ or $r