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\begin{document}
\today

\bigskip
\title{Theta correspondence for unitary groups}

%\author{Dan Barbasch}
%       \address[D. Barbasch]{Dept. of Mathematics\\
%                Cornell University\\Ithaca, NY 14850}
%        \email{barbasch@math.cornell.edu}

\maketitle

\bigskip
\section{Notation}\label{sec:1}

\subsection{}\label{sec:1.1}
For a reductive Lie algebra $\fk g,$ the complexification of a real
Lie algebra  $\fk g_0,$ let $\fk g=\fk k +\fk s$ be the Cartan
decomposition. In the case of $u(p,q),$ we think of the Lie algebra as
$\Hom_\bC[V,V],$ where $V=V^+ + V^-.$ Then the Cartan decomposition is
\begin{equation}
  \label{eq:1.1.1}
  \begin{aligned}
  \fk k=&(\Hom[V^+,V^+] + \Hom[V^-,V^-])\\ 
  \fk s=&(\Hom [V^+,V^-]   +\Hom[V^-,V^+]).  
\end{aligned}
\end{equation}
 The action of $K_c=GL(V^+)\times GL(V^-)$ on $\fk s$  is given by
 \begin{equation}
   \label{eq:1.1.2}
   (g^+,g^-)\cdot (A,B):=(g^+A(g^-)^{-1},g^-B(g^+)^{-1}).
 \end{equation}
A nilpotent orbit is represented by a pair $\eta:=(A,B)$ which satisfies
$AB$ (equivalently $BA$) is nilpotent. We will view $\eta$ as an
element of $\Hom[V,V]$ whenever convenient. It can also be represented by a
partition $(p_1,\dots ,p_k)$ and corresponding Young diagram where
each entry is labelled $\pm$ alternating on the rows.  The signs in the
first column give the decomposition of the kernel of $\eta$ according
to $V^\pm,$ the signs in the first two columns the decomposition of
the kernel of $\eta^2$ according to $V^\pm,$ and so on.

The transpose partition is denoted $(\wht{p}_1,\dots ,\wht{p}_t).$

Write $C_\eta$ for the orbit of $\eta$ under the group $K_c.$ The
following lemma describes closure relations.
\begin{lemma}
  \label{l:1.1}
Let $\eta$ and $\tau$ be nilpotent orbits, and assume the partitions
have the same number of columns by padding one with zeroes if
necessary. Then $C_{\tau}\subset \ovl{C}_{\eta},$ if and only if for
every $i,$ the number of $\pm$ in the largest $i$ columns of $\tau$
is greater than or equal to the corresponding number of $\pm$ for
$\eta.$

Alternatively,  $C_{\tau}\subset \ovl{C}_{\eta}$ if and only if there is a
sequence of $(\eta_i,\eta_{i+1})$ such that 
\begin{itemize}
\item $\eta_1=\eta,\ \eta_n=\tau,$ and
\item $\eta_{i+1}$ is obtained from $\eta_i$ as
follows. Remove a chunk from the beginning (respectively) end of one
row, of $\eta_i,$ and add it to the beginning (end respectively) of the next
smaller row for which the result is still the tableau of a
nilpotent orbit. 
\end{itemize}
\end{lemma}
\begin{proof}
The necessity of the condition follows from the fact that 
\begin{equation}
  \label{eq:1.1.3}
  \dim \ker\tau^i_{V^\pm}\ge \dim \ker\eta^i_{V^\pm}.
\end{equation}
The sufficiency of the condition follows by direct calculation, using the
alternate description of the condition. 
\end{proof}
\subsection{}\label{sec:1.2}
We define a sequence of partitions 
\begin{equation}
  \label{eq:1.2.1}
  \eta_t,\ \dots ,\eta_1,
\end{equation}
where $\eta_t=\eta,$ and $\eta_i$ is obtained from $\eta_{i+1}$ by
removing the largest column. We construct spaces 
\begin{equation}
  \label{eq:1.2.2}
  (U_1^+,U_1^-),\ \dots \ , (U_t^+,U_t^-),
\end{equation}
where each $U_i^\pm$ has dimension the number of $\pm$ in the tableau
of $\eta_i.$ 

Note that in this situation,
\begin{equation}
  \label{eq:1.2.3}
  \begin{aligned}
    &\dim U_i^+\ge (\dim U^-_{i-1}-\dim U^-_{i-2}) +\dim U_{i-1}^+,\\
    &\dim U_i^-\ge (\dim U^+_{i-1}-\dim U^+_{i-2}) +\dim U_{i-1}^-,\\
  \end{aligned}
\end{equation}
\subsection{}\label{sec:1.3}
Write $U_i:=U_i^+ + U_i^-.$ Write $L$ for $\Hom.$ The space
\begin{equation}
  \label{eq:1.3.1}
  \CW:=L(U_1,U_2)\times L(U_2,U_1)\times\dots\times
  L(U_{t-1},U_t)\times L(U_t,U_{t-1})
\end{equation}
is symplectic. We choose a Lagrangian subspace
\begin{equation}
  \label{eq:1.3.2}
  \begin{aligned}
  \C L:= &L(U_1^+,U_2^+)+L(U_2^+,U_1^-)+L(U_1^-,U_2^-) +L(U_2^-,U_1^+)
  + \dots \\
 &+L(U_{t-1}^+,U_t^+)+L(U_t^+,U_{t-1}^-)+L(U_{t-1}^-,U_t^-) + 
   L(U_t^-,U_{t-1}^+)
  \end{aligned}
\end{equation}
A typical element of this space is 
\begin{equation}
  \label{eq:1.3.3}
  \al:=(A_1^+,B_1^+,A_1^-,B_1^-,\dots ,A_{t-1}^+,B_{t-1}^+,
        A_{t-1}^-,B_{t-1}^-)
\end{equation}
\subsection{}\label{sec:1.4}
Let 
\begin{equation}
  \label{eq:1.4.1}
  \C N:= L(U_1^+,U_1^-)\times L(U_1^-,U_1^+)\times \dots
 \times L(U_{t-1}^+,U_{t-1}^-)\times L(U_{t-1}^-,U_{t-1}^+).
\end{equation}
Then the restriction of the map $\Phi$ in \cite{KP} is given by
\begin{equation}
  \label{eq:1.4.2}
  \begin{aligned}
\Phi(\al):=&
(B_1^+ A_1^+,B_1^-A_1^-,B_2^+A_2^+-A_1^-B_1^+,B_2^-A_2^-
-A_1^+B_1^-,\dots ,\\ 
&B_{t-1}^+A_{t-1}^+-A_{t-2}^+B_{t-2}^-, 
 B_{t-1}^-A_{t-1}^--A_{t-2}^-B_{t-2}^+). 
\end{aligned}
\end{equation}
The variety $\C Z$ is defined as $\Phi^{-1}(0).$ The group $K_c$ acts
in the obvious way, extending (\ref{eq:1.1.2}).  Consider the map
\begin{equation}
  \label{eq:1.4.3}
  \begin{aligned}
  &\Theta: \C L\longrightarrow L(U_t^+,U_t^-)\times
  L(U_t^-,U_t^+),\\ 
  &\Theta(\al):=(A_{t-1}^+B_{t-1}^-,A_{t-1}^-B_{t-1}^+).
  \end{aligned}
\end{equation}

\bigskip
\section{The variety $\C Z$}\label{sec:2}

\subsection{}\label{sec:2.1} We consider the case when there are just
two vector spaces $U_1=U_1^+ + U_1^-$ and $U_2=U_2^+ + U_2^-.$ An
element  $\ep\in \C L$ can be viewed as a \textit{cyclic quiver} 
\begin{equation}
  \label{eq:2.1.1}
  U_1^+\ovr{A_1^+}{\longrightarrow} U_2^+\ovr{B_1^+}{\longrightarrow}
  U_1^-\ovr{A_1^-}{\longrightarrow}  U_2^-\ovr{B_1^-}\longrightarrow U_1^+,
\end{equation}
and we can talk about the powers $\ep^i.$  
The orbit structure of $\C L$ under the group $K_c$ is given by 
partitions/tableaus $T$ where each row is labelled 
\[
\boxed{\dots |a|b|c|d|a|\dots }\ .
\]
The number of $a$'s in the largest $i$ rows is the dimension of
the kernel of $\ep^i$ on $U_1^+,$ and so on.

This is the analogue of the Jordan canonical form for a single
nilpotent transformation.

The pairs 
\begin{equation}
  \label{eq:2.2.1a}
  \begin{aligned}
&(B_1^+A_1^+,B_1^-A_1^-)\in L(U_1^+,U_1^-)\times L(U_1^-,U_1^+),\\
&(A_1^+B_1^-,A_1^-B_1^+)\in L(U_2^+,U_2^-)\times L(U_2^-,U_2^+)   
  \end{aligned}
\end{equation}
correspond to tableaus $T_{ac}$, and $T_{bd}$ respectively, obtained
from $T$ by removing the $b,d$'s, and $a,c$'s respectively.


\subsection{}\label{sec:2.2}
In view of (\ref{eq:1.2.3}), we consider the special case when

\begin{equation}
  \label{eq:2.2.1}
\dim U_1^+ + \dim U_1^-\le \dim U_2^+,\ \dim U_2^- .
\end{equation}

Let $T\in \C L,$ and write
$\eta_1:=\Phi_1(T),\ \eta_2:=\Phi_2(T).$ Then 
\begin{equation}
  \label{eq:2.2.2}
\eta_1=(B_1^+A_1^+,B_1^-A_1^-),\qquad 
\eta_2=(B_2^+A_2^+-A_1^-B_1^+,B_2^-A_2^--A_1^+B_1^+). 
\end{equation}
The tableau corresponding to $\eta_1$ is denoted 
$T_{bd},$ the one corresponding to $\eta_2$, is denoted
$T_{ac}.$ They are obtained by removing $b,c$ or $a,d$ respectively
from $T.$ The correspondence with the $\pm$ tableaus is $a,d\mapsto
+,$ and $b,c\mapsto -.$  


\begin{proposition}
  \label{p:1.4}
The image of $\C Z$ under $\Theta$ is  $\ovl{C}_\eta,$ the closure of the
orbit of $\eta.$ 
\end{proposition}
\begin{proof}
We do an induction on $t.$ For $t=1,$ consider the case
$(A_1^+,B_1^+,A_1^-,B_1^-).$ Let $T_{bd}$ correspond to the case 
$(B_1^+ A_1^+,B_1^-A_1^-)=(0,0).$ It follows 
that $T_{ac}$ is a single column with the number of $a$'s equal to
$\dim U_1^+,$ and number of $c$'s equal to $\dim U_1^-.$ 
Let  $T$ map to this $T_{ac}.$ It is obtained from $T_{ac}$ by 
\begin{enumerate}
\item  adding $b$'s to the beginning of some rows starting with $c,$
  or the the end of rows finishing in $a,$
\item adding $d$'s to the beginning of some rows starting with $a,$
  or the the end of rows finishing in $c,$
\item adding the rest of the $b,d$'s as rows each formed of a single $b,$
  and rows each formed of a single   $d$ at the bottom of the tableau. 
\end{enumerate}
Let $T^{max}$ be the tableau obtaind from $T_{ac}$ by replacing each
row equal to $a$ by $dab,$ each row of the form $c$ by $bcd,$ and
adding the rest of the $b$'s and $d$'s as single rows at the bottom. 
This is possible because of (\ref{eq:2.2.1}). Then lemma \ref{l:1.1}
implies that the orbit given by  $T_{bd}$ corresponding to
$T^{max}$ contains all others in its closure. Thus  $(A_1^-B_1^+,A_1^+B_1^-)$ is in
$\ovl{C}_{\eta_2}.$   

%%%%%%%%%%%%%%%%%%%%%%%


Now assume by induction that
$(A_{i-1}^-B_{i-1}^+,A_{i-1}^+B_{i-1}^-)\in \ovl{C}_{\eta_i}.$ Let $T$
map to a $T_{ac}$ corresponding to this. Then 
\begin{equation}
  \label{eq:1.4.4}
 (B_i^+A_i^+,B_i^-A_i^-)=(A_{i-1}^-B_{i-1}^+,A_{i-1}^+B_{i-1}^-)\in
 \ovl{C}_{\eta_i}.
\end{equation}
The tableau $T$ is obtained from $T_{ac}$ by 
\begin{enumerate}
\item inserting a $b$ between every $ac,$ and a $d$ between every $ca,$
\item adding $b$'s to \textbf{some} rows starting with $c,$ and some rows
  ending in $a,$ 
\item adding $d$'s to \textbf{some} rows starting with $a,$ and some rows
  ending in $c,$
\item adding the remainig $a$'s and $c$'s at the bottom as single rows.
\end{enumerate}
Fix a $T_{ac}.$ As before we can construct a $T^{max}$ by applying
(1)-(4) above, but in (2), and (3), adding $b,d$ to the beginning and
end of \textbf{all} rows.
Let $\eta^{max}$ be the orbit corresponding to $T^{max}_{bd}.$ 
By lemma \ref{l:1.1} any other tableau $T_{bd}$ obtained from  $T$ mapping to
the fixed $T_{ac}$ corresponds to an orbit in the closure of the one
for $eta^{max}.$ In turn the largest $\eta^{max}$ corresponds to $\eta_{i+1}.$
This means,
\begin{equation}
  \label{eq:1.4.5}
  (A_{i}^-B_{i}^+,A_{i}^+B_{i}^-)\in \ovl{C}_{\eta_{i+1}}.
\end{equation}
\end{proof}

\subsection{}\label{sec:2.2a}
\begin{proposition}
  \label{p:2.2a}
The dimension of the orbit of $\ep$ equals the sum of
\begin{enumerate}
\item for each row, $\boxed{x|\dots |y},$ the number of $x$ in the
  later/smaller   rows plus the number of $y$ in the later/smaller
  rows. If a row has   size 1, then count twice,
\item for each row $\boxed{x|\dots |y},$ the number of $x$ in
  the row itself. 
\end{enumerate}
\end{proposition}
\begin{proof}
Let $U=U_1^+ +U_2^+ + U_1^- + U_2^-, $ and view $\ep$ as an
endomorphism of $U.$ Let $\zeta$ be a primitive
fourth root of unity, and define a diagonal element 
\begin{equation}
  \label{eq:2.1.2}
  \rho:=(1,\dots ,1,\zeta,\dots ,\zeta,\zeta^2,\dots ,
  \zeta^2,\zeta^3,\dots ,\zeta^3)
\end{equation}
such that its centralizer is $K_c.$ Then 
\begin{equation}
  \label{eq:2.1.3}
  [\rho,\ep]=\zeta\ep.
\end{equation}
Conjugate $\ep$ so that it is in Jordan canonical form, with the
blocks in decreasing order, and $\rho$ is still diagonal. Then on each
block, $\rho$ is of the form
\begin{equation}
  \label{eq:2.1.4}
  (r,r\zeta,r\zeta^2,r\zeta^3,r,\dots ).
\end{equation}
If $r=1,$ the corresponding row in the tableau starts with $a,$ if 
$r=\zeta$ it starts with $b,$ and so on. Now $\rho$ acts on the
centralizer of $\ep,$ and the centralizer in $\fk k_c$ is the fixed
space under $\rho.$ 

Write 
\begin{equation}
  \label{eq:2.1.5}
  U=W_1+\dots +W_k,
\end{equation}
for the block decomposition of $\ep.$ Suppose $i<j$ and therefore
$\dim W_i\ge \dim W_j.$ 
Then the contributions of  $\Hom[W_i,W_j]$ and $\Hom[W_j,W_i]$ to the
centralizer are given by the matrices
\begin{equation}
  \label{eq:2.1.6}
  \begin{aligned}
  &\begin{bmatrix}
    I\\0
  \end{bmatrix},\qquad \begin{bmatrix}
   0&I 
  \end{bmatrix}\\ 
&I=  \begin{bmatrix}
  x_1&x_2&\dots &x_{\ell-1}&x_\ell\\
  0  &\ddots&\ddots&\ddots &x_{\ell-1}\\
  \vdots&&\ddots &\ddots &\vdots\\
  0&&&\ddots&x_2 \\
  0&&\dots &\dots &x_1\\
    \end{bmatrix}\\
  \end{aligned}
  \end{equation}
The first sum comes from counting the off-diagonals centralized by
$\rho.$ 
For $i=j,$ the same argument applies, but there is only one
$\Hom[W_i,W_i].$ 

\end{proof}
\subsection{}\label{sec:2.3}
We fix $T_{bd},$ and consider all the $T\in\C L$ that map to
it. 
\begin{definition}
We call $T_{max}$ the tableau obtained from $T_{bd}$ as follows:

\textit{Insert $a,c$ between all the $b,d,$ and add one $a,c$ to the beginning
and to the end of each row. If there are any $a,c$ left, add them as
single rows to the bottom.}
\end{definition}

\begin{proposition}
  \label{p:2.3}
Suppose $T$ maps to $T_{bd}.$ Then $\dim C_T\le \dim C_{T_{max}}.$
\end{proposition}
\begin{proof}
Any such tableau is obtained from $T_{max}$ by moving some $a,c$ from
the beginning or end of rows, and moving them down to the bottom of the
tableau. Then in fact 
\begin{equation}
  \label{eq:2.3.1}
  C_T\subset \ov{C_{T_{max}}}. 
\end{equation}
We also check the dimension statement directly.
Number the rows in decreasing order, so that the largest one
is the first (equal size rows in arbitrary order). Then the number of
$a$ and $c$ in the rows higher than $i$ can only increase compared to
the corresponding number for $T_{max}.$ Consider the number
$n_i$ contributed by row $i.$
\begin{itemize}
\item If row $i$  starts with $a,c$, then $n_i$ can only
  incresase. Similarly if it ends in $a$ or $c.$
\item Suppose row $i$ for $T$ starts with $b$ ($d$ is similar). 
  \begin{itemize}
\item Each $b$ but the first one in row $i$ is preceded by an $a.$
  Thus the number for $i$ is either the same as for $T_{max}$, or has
  decreased by 1. 
\item The number of $b$ in the lower rows of $T$ are at most one less
  than the number of $a$'s in the corresponding row of $T_{max}.$  
\item Each row of $T$ later than $i$ that has fewer $b$'s than
     $a$'s must end in an $a$ in $T_{max}.$ Thus there is an extra
     count of 1 from the $a$ that has moved from row $i.$ This offsets
     the drop of 1 from counting the $b$'s.  
\item The $a$ that has moved down contributes another 1. 
  \end{itemize}
 
\end{itemize}
\end{proof}
\subsection{}\label{sec:2.4}
\begin{proposition}
Assume that $\Phi$ is such that $\Phi_{bd}$ satisfies
$C_{\Phi_{bd}}\subset\ov{C_{T_{bd}}}.$ Then
  \begin{equation*}
    \dim C_\Phi\le \dim C_{T_{max}}.
  \end{equation*}
\end{proposition}
\begin{proof}
  It is enough to consider the case when $\Phi_{bd}$ is obtained from
  $T_{bd}$ by removing the edge (beginning or end) of one row and
  moving it down to a lower row. Given such a $\Phi_{bd},$ we can
  construct $\Phi_{max}$ the same way as we constructed $T_{max}$ from
  $T_{bd}.$ By proposition \ref{p:2.3}, $\dim C_\Phi\le \dim
  C_{\Phi_{max}}.$ Then $\Phi_{max}$ is obtained from $T_{max}$ by moving
   the corresponding edge lower. Thus $\Phi_{max}\subset
  \ov{C_{T_{max}}},$ and the claim follows.
\end{proof}

\subsection{}\label{sec:2.6}
We prove the analogue of lemma \ref{l:1.1} for the
case of cyclic quivers as in (\ref{eq:2.1.1}). Undoubtedly a general
version is known.

Given a tableau $T,$ denote by $A_i,B_i,C_i,D_i$ the number of
$a,b,c,d$ respectively, in the largest $i$ rows. We say $T'\le T,$ if
$T$ can be obtained from $T'$ by applying the following procedure
repeatedly: 
\begin{equation}\label{*} 
\begin{aligned}
&\textit{
Remove a chunk from the beginning (or end) of a row,}\\
&\textit{ and add it to the beginning (or end) of a smaller row.}
\end{aligned}
\end{equation}

\begin{theorem}
$C_T\subset \ovl{C_{T'}}$  if and only if 
$$
A_i\ge A_i',\qquad B_i\ge B_i',\qquad C_i\ge C_i',\qquad D_i\ge
D_i',\quad \text{ for all }\quad  i.
$$  
\end{theorem}
The condition is clearly necessary, so we will concentrate on the
sufficiency. The proof will be given over several sections. It is by
induction on
\begin{equation*}
  D(T,T'):=\sum (A'_i-A_i)+(B'_i-B_i)+(C'_i-C_i)+(D'_i-D_i).
\end{equation*}
If $D(T,T')=0,$ then $T'=T,$ and there is nothing to prove. Otherwise
we will find $T''$ so that $D(T,T'')<D(T,T')$, and $T'$ is obtained
from $T''$ by applying (\ref{*}). Section \ref{sec:2.6} shows that if
$T'$ and $T$ are related by (\ref{*}), then indeed $C_{T'}\subset
\ovl{C_T}.$ sections \ref{sec:2.6}-\ref{sec:2.11} are devoted to
constructing $T''.$

\medskip
We first observe that for all $i,$
\begin{equation}
  \label{eq:2.6.1}
  \begin{aligned}
&A_{i+1}-A_{i}\le D_i-D_{i-1},\\
&B_{i+1}-B_{i}\le C_i-C_{i-1},\\
&C_{i+1}-C_{i}\le B_i-B_{i-1},\\
&D_{i+1}-D_{i}\le A_i-A_{i-1}.
  \end{aligned}
\end{equation} 
This is just the fact that in
each row, a $d$ must be  followed by an $a,$ and so on. 

\subsection{}\label{sec:2.7}
We consider the case when $\ep$ has a $T$ which is formed of just two
rows. We choose a Jordan canonical basis, and write the corresponding
maps as 
\begin{equation}
  \label{eq:2.7.1}
  \begin{aligned}
&x_1\mapsto y_1\mapsto z_1\mapsto w_1\mapsto x_2\mapsto\dots\mapsto
x_m\mapsto y_m\mapsto z_m\mapsto w_m\mapsto 0\\
&\al_1\mapsto\beta_1\mapsto\gamma_1\mapsto\delta_1\mapsto\dots
\mapsto\rho_n\mapsto\tau_n\mapsto\eta_n\mapsto\mu_n\mapsto 0,
\end{aligned}
\end{equation}
where $x_1,\dots $ is a basis of $U_1,$ and $\al_1,\dots $ a basis of
$U_2.$ As in section \ref{sec:2.2a}, let $\rho$ be the semisimple
element so that $[\rho,\ep]=\zeta\ep.$ The notation $x_i,y_i\dots $
reflects the eigenvalue of $\rho$ on the corresponding basis element. 

Let 
\begin{equation}
  \label{eq:2.7.2}
  x=
  \begin{bmatrix}
    I&X\\0&I
  \end{bmatrix}
\end{equation}
be an element of $\Hom[U_1+U_2,U_1+U_2].$ It belongs to $K_c$
precisely when $X$ is fixed by $\rho.$  Assume $m\ge n,$ and let $n\le
k\le m.$ Then there exists $x$ such that $xTx^{-1}$ satisfies
\begin{equation}
  \label{eq:2.7.3}
\mu_n\mapsto r_{m-k-1},
\end{equation}
instead of 0, and the rest the same as before. 
There exists a diagonal element $a_t$ such that
\begin{equation}
  \label{eq:2.7.4}
  \lim_{t\to 0}a_t\cdot r_{m-k}=0,\qquad a_t\cdot s_l=s_l,  
\end{equation}
for all other basis elements $s_l.$
Then $\lim_{t\to 0} a_tTa_t^{-1}$ exists, and is a tableau $T'$ as in
(\ref{eq:2.6.1}), where a chunk from the end of the larger row has
been added to the end of a smaller row. If $T'$ is in the $\zeta$
eigenspace of $\rho,$ $x$ can be chosen to  be in $K_c.$ If $T$
consists of more than two rows, choose a Jordan canonical
basis, and apply the above argument to just two of the rows. 

The argument for when a chunk is added from the beginning of the
larger row to the beginning of the smaller row is similar; basically
interchange $m$ and $n.$



%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\subsection{}\label{sec:2.8}
Assume that the first $k$ columns of $T$ and $T'$ are the same. Then
we can remove them from both, and argue on the remainders. Thus assume
that the first column of $T'$ is longer than that of $T,$ and in
particular,
\begin{equation}
  \label{eq:2.8.1}
  A_1'>A_1.
\end{equation}
Let $k$ be such that $A'_i>A_i$ for $i\le k,$ and $A'_{k+1}=A_{k+1}.$
We claim that there is $j\le k$ such that
\begin{equation}
  \label{eq:2.8.2}
  D'_j-D'_{j-1}>A'_{j+1}-A'_{j}.
\end{equation}
If equality were to hold for all $j\le k,$ we would conclude
\begin{equation}
  \label{eq:2.8.3}
  \begin{aligned}
  D'_k=&(D'_k-D'_{k-1})+\dots + D'_1 =\\ &(A'_{k+1}-A'_k)+\dots +
  (A'_2-A'_1)= A'_{k+1}-A'_1<\\ &A_{k+1}-A_1\le D_k,
  \end{aligned}
  \end{equation}
a contradiction.
\subsection{}\label{sec:2.9}
Assume there is an ``a'' in the 1-rows. If there is also a ``d'', then
let $T''$ be the tableau obtained from $T'$ by
\begin{equation}
  \label{eq:2.9.1}
  \begin{matrix}
    a\\d
  \end{matrix}\qquad
\longrightarrow \qquad
\begin{matrix}
  a&d\\
   &
\end{matrix}
\end{equation}
Then $T''$ is between $T$ and $T'$, and we can apply induction.

If there is no ``d'' in  the 1-rows, then
\begin{equation}
  \label{eq:2.9.2}
D'_1=A_2'-A_1'.  
\end{equation}
If a 2-row ends in a ``d'', then make $T''$ by
\begin{equation}
  \label{eq:2.9.3}
  \begin{matrix}
    c&d\\
    a&
  \end{matrix}
\qquad\longrightarrow\qquad
\begin{matrix}
  c&d&a\\
   & &
\end{matrix}
\end{equation} 
Then
\begin{equation}
  \label{eq:2.9.4}
  A''_1=A'_1-1,\qquad A''_2=A'_2-1.
\end{equation}
For $T''$ to be a tableau, it is necessary that $A_2'>A_2.$ But if
$A_2'=A_2,$ then
\begin{equation}
  \label{eq:2.9.5}
  D_1'=A_2'-A_1'=A_2-A_1'<A_2-A_1\le D_1,
\end{equation}
a contradiction. Thus $T''$ is a tableau between $T$ and $T'$ and so
we can apply induction.
Suppose no 1-row or 2-row ends in a ``d'', but a  3-row does. Then
\begin{equation}
  \label{eq:2.9.6}
D'_1=A_2'-A_1',\qquad D'_2-D'_1=A_3'-A'_2.  
\end{equation}
We have already seen that $A'_2>A2.$ Suppose $A'_3=A_3.$ Then
\begin{equation}
  \label{eq:2.9.7}
D'_2=D'_1+(A_3'-A'_2)=A'_3-A'_1=A_3-A'_1<A_3-A_1\le D_2,
\end{equation}
a contradiction. Thus $A'_3>A_3,$ and we can form $T''$ between $T$ and
$T'$ by moving
\begin{equation}
  \label{eq:2.9.8}
  \begin{matrix}
    b&c&d\\
    a& & 
  \end{matrix}
\qquad\longrightarrow\qquad
\begin{matrix}
  b&c&d&a\\
   & & & 
\end{matrix}
\end{equation}
\subsection{}\label{sec:2.10} Assume that $A'_1>A_1,$ and no 1-row of
$T'$ starts with an ``a''. Then
\begin{equation}
  \label{eq:2.10.1}
  B'_2-B'_1=A_1'.
\end{equation}
Thus
\begin{equation}
  \label{eq:2.10.2}
  B'_2=B'_1+A'_1>B'_1+A_1\ge B_1+A_1\ge B_2.
\end{equation}
Assume that $T'$ has a 2-row ``ab''. If one of the
1-rows of $T'$ is a ``d'', then $T''$ obtained by
\begin{equation}
  \label{eq:2.10.3}
   \begin{matrix}
    a&b\\
    d&  
  \end{matrix}
\qquad\longrightarrow\qquad
\begin{matrix}
  d&a&b\\
   & & & 
\end{matrix} 
\end{equation}
is a tableau between $T$ and $T'',$ so we can apply induction. If no
1-rows have a ``d'', then 
\begin{equation}
  \label{eq:2.10.4}
  D'_1=A'_2-A'_1.
\end{equation}
>From (\ref{eq:2.10.2}), $B'_2>B_2,$
and $A'_2>A_2$ (\ref{eq:2.9.4} and \ref{eq:2.9.5}).

Assume that $B'_3=B_3.$ We claim that $A'_2-A'_1>B'_3-B'_2,$ so that
$T'$ has a 2-row ``da''. Indeed, otherwise
\begin{equation}
  \label{eq:2.10.6}
  D'_1=A'_2-A'_1=B'_3-B'_2=B_3-B'_2<B_3-B_2\le A_2-A_1\le D_1,
\end{equation}
a contradiction. Then we can form $T''$ between $T$ and $T'$ by
\begin{equation}
  \label{eq:2.10.6}
   \begin{matrix}
    d&a\\
    a&b\\
  \end{matrix}
\qquad\longrightarrow\qquad
\begin{matrix}
  d&a&b\\
  a& &\\
\end{matrix} 
\end{equation}
and apply induction.

So assume $B'_3>B_3,$ and $T'$ has no 2-row ``da''. Thus
\begin{equation}
  \label{eq:2.10.7}
B'_3-B'_2=A'_2-A'_1.  
\end{equation}
If $B'_4=B_4,$ then
\begin{equation}
  \label{eq:2.10.8}
\begin{aligned}
D'_2-D'_1&=A'_3-A'_2 = B'_4-B'_3=B_4-B'_3\\
&<B_4-B_3\le A_3-A_2\le D_2-D_1\le C_1 
\end{aligned}
\end{equation}
Since also $C_1\le C'_1,$ we conclude that $T'$ has a 1-row ``c'', because
\begin{equation}
  \label{eq:2.10.9}
  D'_2-D'_1< C'_1,
\end{equation}
and also 
\begin{equation}
  \label{eq:2.10.10}
  D'_1>D_1,
\end{equation}
because otherwise
\begin{equation}
  \label{eq:2.10.11}
  D'_2=D'_1 +(D'_2-D'_1)<D_1+(D_2-D_1)=D_2.
\end{equation}
If $T'$ has a 3-row ``dab'', then we can form $T''$ between $T$ and
$T'$ by
\begin{equation}
  \label{eq:2.10.12}
     \begin{matrix}
    d&a&b\\
    c& & 
  \end{matrix}
\qquad\longrightarrow\qquad
\begin{matrix}
  c&d&a&b\\
   & & & 
\end{matrix} 
\end{equation}
If not, then $B'_3-B'_2=C'_4-C'_3.$ Then $C'_4>C_4,$ or else the
analogue of (\ref{eq:2.10.6}) starting with $C'_4-C'_3$ 
implies $D'_1<D_1,$ a contradiction. We can continue this argument
with $D'_5$ and so on, to conclude that we can form $T''$ between $T$
and $T'$ by
\begin{equation}
  \label{eq:2.10.13}
     \begin{matrix}
    d&a&\dots\\
    c& &\\
  \end{matrix}
\qquad\longrightarrow\qquad
\begin{matrix}
  c&d&a&\dots\\
   & & &
\end{matrix} 
\end{equation}

\subsection{}\label{sec:2.11} We continue to assume $A'_1>A_1.$ 
We have dealt with the cases when $T'$
has a 1-row ``a'', and a 2-row ``ab''. Assume that $T'$ has no such
rows, the first row starting with an ``a'' is ``abc''. This implies 
\begin{equation}
  \label{eq:2.11.1}
  \begin{aligned}
    A'_1&=B'_2-B'_1,\qquad \text{ no ``a''},\\
    B'_2-B'_1&=C'_3-C'_2,\qquad \text{ no ``ab''.}
  \end{aligned}
\end{equation}
We conclude
\begin{equation}
  \label{eq:2.11.2}
A'_1>A_1,\qquad B'_2>B_2,\qquad C'_3>C_3.
\end{equation}
Otherwise, if say $C'_3=C_3,$ then
\begin{equation}
  \label{eq:2.11.3}
A'_1=B'_2-B'_1=  C'_3-C'_2\le C_3-C_2\le B_2-B_1\le A_1,
\end{equation}
a contradiction. Thus if $T'$ has a row ``d'', ``da'' or ``dab'', we
can form $T''$ between $T$ and $T'$ by 
\begin{equation}
  \label{eq:2.11.4}
  \begin{aligned}
&\begin{matrix}
    a&b&c\\
    d& & 
  \end{matrix}
\quad\longrightarrow\quad
\begin{matrix}
  d&a&b&c\\
   & & & 
\end{matrix},\\
&\\
&\begin{matrix}
    a&b&c\\
    d&a& 
  \end{matrix}
\quad\longrightarrow\quad
\begin{matrix}
  d&a&b&c\\
  a&& & 
\end{matrix},\\
&\\
&\begin{matrix}
    a&b&c\\
    d&a&b 
  \end{matrix}
\quad\longrightarrow\quad
\begin{matrix}
  d&a&b&c\\
  a&b& & 
\end{matrix}.   
  \end{aligned}
\end{equation}
They give
\begin{equation}
  \label{eq:2.11.5}
  \begin{aligned}
&C''_3=C'_3-1,\qquad &B''_2=B'_2-1,\qquad &A''_1=A'_1-1,\\
&C''_3=C'_3-1,\qquad &B''_2=B'_2-1,\qquad &\\
&C''_3=C'_3-1,\qquad &\qquad &\\    
  \end{aligned}
\end{equation}
which are satisfied by (\ref{eq:2.11.2}). The conclusion follows by
induction. 

\medskip
So assume
\begin{equation}
  \label{eq:2.11.6}
  \begin{aligned}
    D'_1&=A'_2-A'_1,\qquad \text{ no ``d''},\\
    A'_2-A'_1&=B'_3-B'_2,\qquad \text{ no ``da'',}\\
    B'_3-B'_2&=C'_4-C'_3,\qquad \text{ no ``dab''}.    
  \end{aligned}
\end{equation}
As for (\ref{eq:2.11.2}), we conclude that
\begin{equation}
  \label{eq:2.11.7}
A'_2>A_2,\qquad B'_3>B_3,\qquad C'_4>C_4.   
\end{equation}
Otherwise, if say $C'_4=C_4,$ 
\begin{equation}
  \label{eq:2.11.8}
  D'_1=A'_2-A'_1=B'_3-B'_2=C'_4-C'_3<C_4-C_3\le B_3-B_2\le A_2-A_1\le D_1,
\end{equation}
a contradiction (since $D'_1\ge D_1$). We conclude that if $T'$ has a
row ``cd'', ``cda'', ``cdab'', we can form a tableau $T''$ between
$T$ and $T'$ by
\begin{align}  \label{eq:2.11.9}
&\begin{matrix}
    a&b&c\\
    c&d& 
  \end{matrix}
\quad\longrightarrow\quad
\begin{matrix}
  c&d&a&b&c\\
   & & & & 
\end{matrix}\ ,\notag\\
&\notag\\
&\begin{matrix}
    a&b&c\\
    c&d&a 
  \end{matrix}
\quad\longrightarrow\quad
\begin{matrix}
  c&d&a&b&c\\
  a& & & &
\end{matrix}\ ,\\
&\notag\\
&\begin{matrix}
    c&d&a&b\\
    a&b&c 
  \end{matrix}
\quad\longrightarrow\quad
\begin{matrix}
  c&d&a&b&c\\
  a&b& & &
\end{matrix}.\notag   
\end{align}
This gives
\begin{equation}
  \label{eq:2.11.10}
  \begin{aligned}
    &C''_3=C'_3-1,\qquad C''_4=C'_4-1,\\
    &B''_2=B'_2-1,\qquad B''_3=B'_3-1,\\
    &A''_1=A'_1-1,\qquad A''_2=A'_2-1,
  \end{aligned}   
\end{equation}
so (\ref{eq:2.11.2}) and (\ref{eq:2.11.7}) imply that $T''$ is between
$T$ and $T'$. The conclusion follows by induction. 

Thus assume that $T'$ satisfies
\begin{equation}
  \label{eq:2.11.11}
  \begin{aligned}
    D'_2-D'_1&=A'_3-A'_2,\qquad \text{ no ``cd''},\\
    A'_3-A'_2&=B'_4-B'_3,\qquad \text{ no ``cda'',}\\
    B'_4-B'_3&=C'_5-C'_4,\qquad \text{ no ``cdab''}.    
  \end{aligned}
\end{equation}
This implies that 
\begin{align}
    C'_5>C_5,\qquad &B'_4>B_4,\qquad A'_3>A_3,\label{eq:2.11.12}\\
                     &\text{ or }\notag             \\
    D'_2-D'_1<C'_1\qquad &\text{ and }\qquad  D'_1>D_1.\label{eq:2.11.13}
\end{align}
Indeed, if say $C'_5=C_5,$ then 
\begin{equation}
  \label{eq:2.11.14}
  \begin{aligned}
  D'_2-D'_1&=A'_3-A'_2=B'_4-B'_3=C'_5-C'_4<\\
           &C_5-C_4\le B_4-B_3\le  A_3-A_2\le D_2-D_1\le C_1\le C'_1,
  \end{aligned}
\end{equation}
and if $D'_1=D_1,$ we would conclude
\begin{equation}
  \label{eq:2.11.15}
D'_2=(D'_2-D'_1)+D'_1<(D_2-D_1)+D_1=D_2,  
\end{equation}
contradicting the assumption $D'_2\ge D_2.$ 

In case (\ref{eq:2.11.13}), $T'$ has rows
\begin{equation}
  \label{eq:2.11.16}
  \begin{matrix}
    d&a&\dots &x_m\\
    c& &      &
  \end{matrix}\ .
\end{equation}
We can form $T''$ by
\begin{equation}
   \begin{matrix} \label{eq:2.11.17}
    d&a&\dots &x_m\\
    c& &      &
  \end{matrix}
\qquad\longrightarrow\qquad
\begin{matrix}
  c&d&a&\dots &x_m\\
   & & &      &
\end{matrix}
\end{equation}
with  $dab\dots x_m$ the smallest row starting with a ``d''. Then
\begin{equation}
  \label{eq:2.11.18}
  D'_1=A'_2-A'_1=\dots =X'_j-X'_{j-1}=\dots =X'_{m}-X'_{m-1}.
\end{equation}
To have $T''$ between $T$ ad $T',$ we need $X'_j>X_j$ for all $j\le
m.$ As for (\ref{eq:2.11.7}), any equality  $X'_j=X_j$ implies
$D'_1\le D_1,$ a  contradiction. 

In case (\ref{eq:2.11.12}), if $T'$ has any row of the form ``bcd'',
``bcda'' or ``bcdab'', we can form $T''$ by
\begin{align}  \label{eq:2.11.19}
&\begin{matrix}
    b&c&d\\
    a&b&c\\
  \end{matrix}
\quad\longrightarrow\quad
\begin{matrix}
  b&c&d&a&b&c\\
   & & & & &
\end{matrix}\notag\\
&\notag\\
&\begin{matrix}
    b&c&d&a\\
    a&b&c\\
  \end{matrix}
\quad\longrightarrow\quad
\begin{matrix}
  b&c&d&a&b&c\\
  a& & & & &
\end{matrix}\\
&\notag\\
&\begin{matrix}
    b&c&d&a&b\\
    a&b&c& &  
  \end{matrix}
\quad\longrightarrow\quad
\begin{matrix}
  b&c&d&a&b&c\\
  a&b& & & &
\end{matrix}\ .\notag   
\end{align}
Then (\ref{eq:2.11.12}) implies that $T''$ is between $T$ and $T',$
and we can apply induction. 

Thus assume that $T'$ satisfies
\begin{equation}
  \label{eq:2.11.20}
  \begin{aligned}
    D'_3-D'_2&=A'_4-A'_3,\qquad \text{ no ``bcd''},\\
    A'_4-A'_3&=B'_5-B'_4,\qquad \text{ no ``bcda'',}\\
    B'_5-B'_4&=C'_6-C'_5,\qquad \text{ no ``bcdab''}.    
  \end{aligned}
\end{equation}
Suppose 
\begin{equation}
  \label{eq:2.11.21}
C'_6=C_6,\qquad B'_5=B_5,\qquad\text{ or } A'_4=A_4.
\end{equation}
 Then   
\begin{equation}
  \label{eq:2.11.22}
B'_1>D'_3-D'_2\ge 0,\quad \text{ and }\quad D'_2>D_2.  
\end{equation}
Indeed, if say $C'_6=C_6,$
\begin{align}
  D'_3-D'_2=\dots =C'_6-&C'_5<  \label{eq:2.11.23}\\ 
&C_6-C_5\le\dots\le D_3-D_2\le C_2-C_1\le
  B_1\le B_1'.     \notag
\end{align}
Condition $D'_2>D_2,$ comes from the fact that if $D'_2=D_2,$ then
\begin{equation}
  \label{eq:2.11.24}
  D'_3=(D'_3-D'_2)+D'_2<(D_3-D_2)+D_2=D_3.
\end{equation}
Condition $B'_1>D'_3-D'_2$ implies that $T'$ has a row ``b'' or
``bc''. If $C'_2-C'_1=D'_3-D'_2,$ there is no ``bc''. In addition,
$C'_2-C'_1<C_2-C_1,$ so $C'_1>C_1.$ Otherwise
\begin{equation}
  \label{eq:2.11.25}
  C'_2=(C'_2-C'_1)+C'_1<(C_2-C_1)+C_1=C_2,
\end{equation} 
a contradiction. Thus (still assuming (\ref{eq:2.11.21}) ), in
addition to $D'_2>D_2$ and $B'_1>0,$ 
\begin{align}
 &C'_2-C'_1=D'_3-D'_2,\quad\text{ so also }\quad 
C'_1>C_1,\ B'_1>C'_2-C'_1,
\label{eq:2.11.26}\\
 &\text{ or }\notag\\
 &C'_2-C'_1>D'_3-D'_2. \label{eq:2.11.27}
\end{align}
In case (\ref{eq:2.11.26}), $T'$ has rows
\begin{equation}
  \label{eq:2.11.28}
  \begin{matrix}
    c&d&a&\dots &x_m\\
    b& & &      &
  \end{matrix}\ .
\end{equation}
The argument in case (\ref{eq:2.11.13}) applies, so that $T''$
obtained by
\begin{equation}
  \label{eq:2.11.29}
  \begin{matrix}
    c&d&a&b&\dots &x_m\\
    b& & & &      &
  \end{matrix}
\quad\longrightarrow\quad
\begin{matrix}
  b&c&d&a&b&\dots &x_m\\
   & & & & &      &  
\end{matrix}\ 
\end{equation}
is between $T$ and $T'.$ 

\medskip
In case (\ref{eq:2.11.27}), $T'$ has a 2-row ``bc''. Suppose
$D'_1=D_1,$ then $C'_1>0.$ Otherwise $C_1=0$ as well, and 
\begin{equation}
  \label{eq:2.11.30}
  0=C'_1\ge D'_2-D'_1,\qquad 0=C_1\ge D_2-D_1=0.
\end{equation}
Then 
\begin{equation}
  \label{eq:2.11.31}
  D'_2=D'_1=D_1=D_2,
\end{equation}
a contradiction. Thus $T'$ has two rows
\begin{equation}
  \label{eq:2.11.32}
\begin{matrix}
   c&d&a&b&\dots &x_m\\
   b&c& & & &      &  
\end{matrix}\ ,   
\end{equation}
and we can apply the argument for (\ref{eq:2.11.26}).

\medskip
Thus assume (\ref{eq:2.11.27}) and $D'_1>D_1.$ Consider $T''$ obtained
by
\begin{equation}
  \label{eq:2.11.33}
  \begin{matrix}
    d&a&\dots &x_m\\
    b&c&      &
  \end{matrix}
\qquad\longrightarrow\qquad
\begin{matrix}
    b&c&d&a&\dots &x_m\\
     & & & &      &
\end{matrix}\ .
\end{equation}
Then $T''$ satisfies
\begin{equation}
  \label{eq:2.11.34}
  \begin{aligned}
&D''_1=D'_1-1,\dots,X''_i=X'_i-1,\dots ,X''_m=X'_m-1,\\
&D''_2=D'_2-1,\dots,Y''_{j}=Y'_j-1,\dots ,Y''_{m+1}=Y'_{m+1}-1.
  \end{aligned}
\end{equation}
If any $X'_i=X_i,$ then as in (\ref{eq:2.11.18}), we get $D'_1\le D_1,$
a contradiction to the hypothesis. If on the other hand $Y'_j=Y_j,$ we
get (\ref{eq:2.11.14}). Thus  $T''$ is between $T$ and $T'.$ 

\medskip
We have reduced to the case 
\begin{equation}
  \label{eq:2.11.35}
  C'_6>C_6,\qquad B'_5>B_5,\qquad A'_4>A_4.
\end{equation}
\boxed{\textbf{ ???\ The argument continues in a similar fashion.\ ???}}































%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{comment}
If $B'_3>B_3,$  we can form  the tableau $T''$ between $T$ and $T'$ by
\begin{equation}
  \label{eq:2.10.5}
    \begin{matrix}
    a&b\\
    c&d  
  \end{matrix}
\qquad\longrightarrow\qquad
\begin{matrix}
  c&d&a&b\\
   & & & 
\end{matrix}  
\end{equation}
The claim in this case follows by induction.


Suppose $T'$ has no 1-row ``d'', and no 2-row ``cd''. Then
\begin{equation}
  \label{eq:2.10.7}
  D'_1=A'_2-A'_1,\qquad D'_2-D'_1=A'_3-A'_2.
\end{equation}
>From before, if $B'_3=B_3,$ $T'$ must have  a 2-row ``da''. Indeed,
$B'_2>B_2$ (\ref{eq:2.10.2}). If $T'$ has no 2-row ``da'', then
(\ref{eq:2.10.6}) gives a contradiction.  
\end{comment}






%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% 
\newpage
\section{Symplectic and Orthogonal groups}\label{sec:3}  

\subsection{}\label{sec:3.1} We consider the pair $sp(\CR)\times
o(\CS).$ Write $\CR=\C L +\C L^\perp,$ for the symplectic space, and
$\C S=V+W$ for the orthogonal space.  The basis for $\C L$ is denoted by
$e$'s, the dual basis in $\C L^\perp$ by $f.$ A nilpotent element in
$\frak s\cap sp$ is formed of orthogonal blocks. An even sized block
is of the form   
\begin{equation}
  \label{eq:3.1.1}
  \begin{aligned}
  &e_1\mapsto f_2\mapsto e_3\mapsto \dots\mapsto e_n\mapsto f_n\mapsto
  e_{n-1}\dots  \mapsto f_1\mapsto 0,\\
  &e_1\mapsto f_2\mapsto e_3\mapsto \dots\mapsto f_n\mapsto e_n\mapsto
  e_{n-1}\dots  \mapsto f_1\mapsto 0,
  \end{aligned}
\end{equation}
depending whether $n$ is odd or even, or the reverse order. 

An odd sized block has to come in pairs,
\begin{equation}
  \label{eq:3.1.2}
  \begin{aligned}
  &e_1\mapsto f_2\mapsto e_3\mapsto \dots\mapsto e_n\mapsto 0,\\
  &f_n\mapsto e_{n-1}\mapsto \dots\mapsto f_1\mapsto 0,
  \end{aligned}  
\end{equation}
or
\begin{equation}
  \label{eq:3.1.3}
  \begin{aligned}
  &e_1\mapsto f_2\mapsto e_3\mapsto \dots\mapsto f_n\mapsto 0,\\
  &e_n\mapsto e_{n-1}\mapsto \dots\mapsto f_1\mapsto 0,
  \end{aligned}  
\end{equation}
depending whether $n$ is even or odd. Take a pair $(A,B)\in\C Z$
mapping to a nilpotent orbit in $sp(\C R).$ This coincides with the
case of the unitary groups via taking the quadruple
\begin{equation}
  \label{eq:3.1.4}
  A\in\Hom(\C L, V),\ B\in\Hom(\ L^\perp, W),\ A^*\in\Hom(V,\C
  L^\perp),\ B^*\in\Hom(W,\C L).
\end{equation}
We write
\begin{equation}
  \label{eq:3.1.5}
  A:e_i\mapsto x_i,\qquad B':f_j\mapsto y_i,
\end{equation}
with $x_i\in V,\ y_i\in W.$
Then for an even block, we get
\begin{equation}
  \label{eq:3.1.6}
  e_1\mapsto x_1\mapsto f_2\mapsto y_2\dots \mapsto f_1\mapsto
  y_1\mapsto 0,
\end{equation}
or the reverse. In case $n$ is even, we find that
$|x_{2i-1}|=|x_{2i}|=0,$ and $(x_{2i-1},x_{2i})=1,$ while $|x_n|=1.$
The rest of the inner products are 0. Similarly,
$|y_{2i}|=|y_{2i+1}|=0,$ $(y_{2i},y_{2i+1})=1,$ the rest of the inner
products are 0.  There are two cases, $y_1=0$ and $y_1\ne 0,$ and in
either situation, $|y_1|=0.$ In the
first case, the string is (\ref{eq:3.1.6}), and we write it as
\begin{equation}
  \label{eq:3.1.7}
  abcd\dots c,\qquad \text{ or }\qquad cdab\dots a
\end{equation}
where (contrary to upq) the last entry on the right represents the
kernel. When $y_1\ne 0,$ there is a dual vector $y_1',$ which maps to
$e_1.$ This corresponds to the string
\begin{equation}
  \label{eq:3.1.8}
  dabc\dots d,\qquad\text{ or }\qquad  bcda\dots b.
\end{equation}
For the odd blocks, we get pairs of strings
\begin{equation}
  \label{eq:3.1.9}
  \begin{aligned}
    &\begin{matrix}a&b&c&d&\dots&a\\ 
                   c&d&a&b&\dots&c\\
     \end{matrix} ,\\
     &\\
    &\begin{matrix}
    a&b&c&d&\dots&a&\boxed{b}\\
    \boxed{b}&c&d&a&\dots&d&a
    \end{matrix}\\
    &\\
    &\begin{matrix}
    \boxed{d}&a&b&c&\dots&d&a\\
    c&d&a&b&\dots&c&\boxed{d}
    \end{matrix}\\
    &\\
    &\begin{matrix}
    \boxed{d}&a&b&c&\dots&a&\boxed{b}\\
    \boxed{b}&c&d&a&\dots& c&\boxed{d}
    \end{matrix}
  \end{aligned}
\end{equation}



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\end{document}
