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\begin{document}
\title[Singular loci of varieties of complexes]
{SINGULAR LOCI OF VARIETIES OF COMPLEXES}
\author[V. Lakshmibai]{V. Lakshmibai${}^{\dag}$}
\thanks{${}^{\dag}$Partially supported by NSF Grant DMS 9502942.}
\maketitle
%%%%%%%%%%%%%
\section*{Introduction}
Let $V_1,\ldots, V_{h+1}$ be vector spaces
over an arbitrary field $K$
with respective dimensions $n_1,\ldots,n_{h+1}$. Let $Z$ be the affine space of all
$h$-tuples of linear maps $(f_1,\ldots,f_{h}):V_1 \stackrel{f_1}{\to} V_2
 \stackrel{f_2}{\to} \cdots \stackrel{f_{h\!-\!1}}{\to} V_{h}
 \stackrel{f_{h}}{\to} V_{h+1} \ .$
If we endow each $V_i$ with a basis, we get $V_i \cong K^{n_i}$
and $
Z \cong M(n_2 \!\times\! n_1) \!\times\! \cdots
\times M(n_{h+1} \!\times\! n_{h}) ,$
where $M(l\!\times\! m)$ denotes the affine space of matrices
over $K$ with $l$ rows and $m$ columns. Let $V$ be the variety of complexes, namely, the
subvariety of
$Z$ consisting of
$\{(A_1,A_2,\cdots ,A_h)\ |\ A_iA_{i-1}=0, \ 2\le i\le h\}$. It is shown in
\cite{mu-se} (see also \cite{l-m}) that each irreducible component of $V$ is
isomorphic to the {\it opposite cell} in a Schubert variety in $G/Q$, where
$G=SL(n),\ n = n_1  + \cdots + n_{h+1}$, and $Q$ is a certain parabolic sub group.
Let $C$ be an irreducible component of $V$. In this paper, we determine the singular
locus of $C$ (cf. Theorem \ref{5.2}), for the case $h=2$. Let $X(w)\, (\subset G/B)$
be the Schubert variety associated to $C$. We further show (Theorem \ref{7.3}) that
the conjecture of
\cite{l-s} holds for $X(w)$. These $w$'s turn out to be non-vexillary (cf.
\cite{l}), i.e., the corresponding permutations involve the pattern $c,d,a,b$, where
$a<b<c<d$.


\section{Quiver varieties.}
Let $V_i,\ 1\le i \le h+1$, $Z$ be as in the Introduction. Let
$\bold{n} = (n_1,\ldots,n_{h+1})$.
The group $
G_{\bold{n}} = GL(n_1) \times \cdots \times GL(n_{h+1})$
acts on $Z$ by
$$
(g_1,g_2,\cdots,g_{h+1}) \cdot (f_1,f_2,\cdots,f_{h})
= (g_2 f_1 g_1^{-1}, g_3 f_2 g_2^{-1},\cdots,
g_{h+1}f_{h} g_{h}^{-1}).
$$


Now, let $\bold{r} = (r_{ij})_{1 \leq i \leq j \leq h+1}$
be an array of non-negative integers with $r_{ii} = n_i$;
and define $r_{ij} = 0$ for any indices other than
$1\leq i\leq j \leq h+1$.  Define the sets
$$
Z^{\circ}(\bold{r}) = \{(f_1,\cdots,f_{h}) \in Z
\ \mid\  \forall\, i\!<\!j,\ \text{ rank } (f_{j\!-\!1} \cdots f_i : V_i \to V_j)
 = r_{ij} \},
$$
 $$Z(\bold{r})={\overline{Z^{\circ}(\bold{r})}} .$$
(These sets might be empty for a bad choice of $\bold{r}$.)
\\[1em]
\begin{prop}(see \cite{l-m} for example)
\begin{enumerate}
\item The $G_{\bold{n}}$-orbits of
$Z$ are exactly the sets $Z^{\circ}(\bold{r})$
for $\bold{r}=(r_{ij})$ with $
r_{ij}-r_{i,j\!+\!1}-r_{i\!-\!1,j}+r_{i\!-\!1,j\!+\!1}
\geq 0,\quad
\forall\ 1\! \leq\! i\! <\! j\! \leq\! h+1.$
\item  $Z(\bold{r})= \{(f_1,\cdots,f_{h}) \in Z
\ \mid\  \forall\, i\!<\!j,\ \text{ rank } (f_{j\!-\!1} \cdots f_i : V_i \to V_j)
 \le r_{ij} \}$.
\end{enumerate}

\end{prop}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{The Schubert variety $X_Q(w)$.}

Given $\bold{n}=(n_1,\cdots,n_h)$, for $1 \leq i \leq h+1$, let $ a_i = n_1  + \cdots +n_i, \  a_0 = 0,\
n = n_1  + \cdots + n_{h+1} \ .$
For positive integers $i \leq j$, we shall frequently use
the notations
$$
[i,j] = \{ i, i+1, \ldots, j\}, \ [i] = \{i\},
\ [0] = \{\},\text{ the empty set} \ .
$$
  Consider $GL(n)$, its subgroup $B$ of upper-triangular
matrices, and the parabolic subgroup $
Q = \{ (a_{ij}) \in GL(n) \mid a_{ij}=0 \
\mbox{whenever}\ j\leq a_k <i
\ \mbox{for some}\ k \}\ .$
We have, $Q=P_{a_1}\cap \cdots \cap P_{a_h}$, where for $1\le i\le n-1,\ P_i $ denotes the maximal parabolic sub
group associated to the simple root $\a_i$, the simple roots being indexed as in \cite{bou}.
\subsection{The partially ordered set $I_{a_1,\dots,a_h}$}
 Let
$$I_{a_1,\dots,a_h}=\{(\ui_1,\dots,\ui_h)\in I_{a_1,n}\times\dots\times
I_{a_h,n}\mid
\ui_t\subset \ui_{t+1}\text{ for all } 1\le t\le h-1\},$$
where for $d<n$, $I_{d,n}:=\{\underline{i}=(i_1,\dots,i_d)\mid 1\le i_1<\dots <i_d\le n\}.$
Then it is easily seen that $W^Q$, the set of minimal representatives of $W/W_Q$
($W$ being the symmetric group $S_n$) may be identified with
$I_{a_1,\dots, a_k}$.

The partial order on the set of Schubert varieties in $G/Q$ (given by inclusion)  induces a
partial  order $\ge$ on $I_{a_1,\dots,a_h}$, namely, for
$\bold{i}=(\ui_1,\dots,\ui_h)$,
$\bold{j}=(\uj_1,\dots,\uj_h)\in I_{a_1,\dots,a_h}$,
$\bold{i}\ge\bold{j}\iff\ui_t\ge\uj_t$  for all  $1\le t\le h$.

\subsection{The opposite big cell in $G/Q$}\label{13.4}
Let $Q=\cap_{t=1}^hP_{a_t}$ be as above. Let $R$ (resp. $R_Q$) denote the root
system of $G$ (resp. $Q$). Denote by
$O^-$  the subgroup of
$G$ generated by
$\{U_\a\mid \a\in R^-\setminus R_Q^-\}$ (here, for a root $\a,\ U_\a$ denotes the $1$-dimensional unipotent
subgroup of
$G$ associated to $\a$). Then
$O^-$ consists of the elements of
$G$ of the form
$$
\begin{pmatrix}
I_1&0 &0 &\cdots&0 &0\\
\ast &I_2&0&\cdots& 0&0\\
\vdots&\vdots&\vdots&&\vdots&\vdots\\
\ast &\ast &\ast &\cdots&I_h&0\\
\ast &\ast &\ast &\cdots&\ast& I_a
\end{pmatrix},
$$
where $I_t$ is the identity matrix of size $a_t-a_{t-1}$, $1\le t\le h$, $I_a$ is the
 identity matrix of size $a$, where $a=n-a_h$, and if
$x_{ml}\ne 0$, with $m\ne l$, then $m>a_t$, $l\le a_t$ for some $t$, $1\le t\le h$.
Further, the restriction of the canonical morphism $f:G\to G/Q$ to $O^-$ is an open
immersion, and
$f(O^-)\simeq B^-e_{\text{id},Q}$. Thus  $B^-e_{\text{id},Q}$ ({\it the opposite
big cell} in $G/Q$) gets identified with
$O^-$.


We shall index the Schubert varieties in $G/Q$ by $W^Q$. For $\t\in W^Q$, we shall
denote the associated Schubert variety by $X_Q(\t)$; we set $Y_Q(\t)=X_Q(\t)\cap O^-$,
{\it the opposite cell } in $X_Q(\t)$ . Note that $Y_Q(\t)$ is a non-empty closed
subvariety of $O^-$.

\subsection{The variety $X_Q(w)$}  Given
$\bold{r} = (r_{ij})_{1\leq i\leq j\leq h+1}$, define $w\in W^Q$ by
$$
w^{(a_i)} = \{\,
\underbrace{1\ldots a_{i\!-\!1}}_
{\mbox{\small $a_{i\!-\!1}$}}
\ \underbrace{. \ldots\ldots a_{i}}_
{\mbox{\small $r_{ii}\!-\!r_{i,i+1}$}}
\ \underbrace{.\ldots\ldots a_{i+1}}_
{\mbox{\small $r_{i,i+1}\!-\!r_{i,i+2}$}}
\ \underbrace{.\ldots\ldots a_{i+2}}_
{\mbox{\small $r_{i,i+2}\!-\!r_{i,i+3}$}}\ \ldots\
\ \underbrace{.\ldots\ldots n_{\mbox{}}}_
{\mbox{\small $r_{i,h+1}$}}
\}
$$
where we use the visual notation $
\underbrace{\cdots\cdots a}_{\mbox{\small $b$}} =
[a\!-\!b\!+\!1,a]$.
\subsection{The map $f$.} Let $z=(A_1,A_2,\cdots ,A_h) \in Z$. Define $f:Z\rightarrow
\cal{O}^{-}
$ by
$$f(z)=\begin{pmatrix}
I_1&0 &0 &0 &\cdots\\
A_1 &I_2&0&0 &\cdots\\
A_2A_1 &A_2 &I_3 &0 &\cdots \\
A_3A_2A_1 &A_3A_2 &A_3 &I_4  &\cdots \\
\vdots&\vdots&\vdots&&\vdots&\vdots
\end{pmatrix}\ (\text{ mod }Q)\ .$$
Denote a generic element of
$ Z = M(n_2\times n_1) \times \cdots
\times M(n_{h+1}\times n_{h})$
by $(A_1,\ldots,A_{h})$, so that the coordinate ring
of $Z$ is the polynomial ring in the entries of all the matrices
$A_i$. Let $\cal{J}(\bold{r}) \subset K[Z]$ be the ideal generated by
the determinantal conditions implied by the definition
of $Z(\bold{r})$:
$$
\cal{J}(\bold{r}) = \left\langle \det(A_{j-1} A_{j-2} \cdots A_i)_
{\lambda\times\mu}
\ \left| \
\begin{array}{c}
j>i,\ \lambda \subset [n_j],\ \mu \subset [n_i] \\[.2em]
\#\lambda = \#\mu = r_{ij}+1
\end{array}
\right.
\right\rangle\ .
$$

\begin{thm}\label{2.4}(cf.\cite{l-m})
\begin{enumerate}
\item $\cal{J}(\bold{r})$ is a prime ideal and is the vanishing ideal
of $Z(\bold{r})\subset Z$.
\item The restriction $f\, |\, _{Z(\bold{r})}$ defines an isomorphism of
$Z(\bold{r})$ onto $Y_Q(w)$.
\end{enumerate}
\end{thm}



%%%%%%%%%%%%%%%%%%
\section{Variety of complexes.}
Let $V$ be the variety of complexes, namely, the
subvariety of
$Z$ consisting of
$\{(A_1,A_2,\cdots ,A_h)\ |\ A_iA_{i-1}=0, \ 2\le i\le h\}$. We have $V=\cup
_{\bold{r}}\ Z(\bold{r}) $, where $r_{ij}=0,\ j\ne i,i+1$. Let us denote $k_i= r_{i
i+1},\
\bold{k}=(k_1,k_2,\cdots ,k_h)$, and $V(\bold{k})=Z(\bold{r})$
. Let
$$A_{\bold{n}}=\{\bold{k}=(k_1,k_2,\cdots ,k_h)\ |\ k_i\le \text{ min
}\{n_i, n_{i+1}\},\ 1\le i\le h,\ k_{i-1}+ k_i \le n_i,\ 2\le i\le h \}.$$ Then
$V=\cup _{\bold{k} \in A_{\bold{n}} }\ V(\bold{k})$.

\subsection{A partial order on $\{(k_1,k_2,\cdots ,k_h)\}$.}\label{3.1} The partial
order on  the set $\{V(\bold{k}),\, \bold{k} \in A_{\bold{n}} \}$ of
$G_{\bold{n}}$-orbit closures in $V$ given by inclusion induces a partial order $\ge
$ on $A_{\bold{n}}$, namely, for
$\bold{k}=(k_1,k_2,\cdots ,k_h),\ \bold{k}'=(k'_1,k'_2,\cdots ,k'_h)$ in
$A_{\bold{n}}$, $\bold{k}\ge \bold{k}'\iff k_t\ge k'_t,\ 1\le t \le h.$
\begin{thm}\label{3.2}
\begin{enumerate}
\item $V(\bold{k})\simeq Y_Q (w)$
\item  $\{V(\bold{k})\ |\ \bold{k}
\text{ is a maximal element of  }A_{\bold{n}} \}$ gives the set of all irreducible
components of $V$.
\end{enumerate}
\end{thm}
\begin{pf}
Assertion (1) follows from Theorem \ref{2.4}. Assertion (2) follows from the fact that
$Z(\bold{r} ')\subset Z(\bold{r})\ \iff r'_{ij}\le r_{ij}, \ 1 \le i<j\le h+1$.
\end{pf}
\begin{rem}
The results in Theorem \ref{3.2} are also proved in \cite{mu-se}.
\end{rem}

Let $Y$ denote the affine space (considered as a subvariety of $O^-$) given by
$$Y=\left\{\begin{pmatrix}
I_1&0 &0 &0 &\cdots\\
A_1 &I_2&0&0 &\cdots\\
0 &A_2 &I_3 &0 &\cdots \\
0 &0 &A_3 &I_4  &\cdots \\
\vdots&\vdots&\vdots&\vdots&\vdots
\end{pmatrix}\right\}.$$ For the rest of the paper we shall identify
$V$ with the subvariety of  $Y$ consisting of $\{(A_1,A_2,\cdots ,A_h)\ |\
A_iA_{i-1}=0, \ 2\le i\le h\}$.

\begin{lem}(cf. \cite{l-m}, \cite{mu-se}) Let $\bold{k} \in A_{\bold{n}}$.
\begin{enumerate}
\item dim $V(\bold{k})=\sum_{1\le i \le h+1}\ (n_i-k_i)\, (k_{i-1}+k_i)$,
where
$k_0=k_{h+1}=0$.
\item codim$_Y V(\bold{k})=\sum_{i=1}^h\ c_i+ \sum_{i=1}^{h-1}\ k_ik_{i+1}$, where
$c_i=(n_{i+1}-k_{i})\,(n_{i}-k_{i})$.
\end{enumerate}
\end{lem}

\section{Singular locus of $V(k_1,k_2)$.} In this section, we take $h=2$, and
determine the singular locus of $V(\bold{k})$. Given $\bold{n}=(n_1,n_2,n_3)$, we fix
 $(k_1,k_2)$ such that $k_i\le \text{ min }\{n_i,n_{i+1}\},\ i=1,2,\ k_1+k_{2}\le
n_2$. For $\bold{r}=(k_1,k_2)$, we shall denote $O(k_1,k_2)=Z^0(\bold{r}),\
V(k_1,k_2)=Z(\bold{r})(=V(\bold{k}))$.
\begin{lem}\label{4.1} Let $c_i=(n_{i+1}-k_{i})\,(n_{i}-k_{i}),\ i=1,2$. We have
\begin{enumerate}
\item $c_1+ k_1k_2-n_1k_2=(n_1-k_1)(n_2-k_1-k_2)$.
\item $c_2+ k_1k_2-n_3k_1=(n_3-k_2)(n_2-k_1-k_2)$.
\end{enumerate}
\end{lem}
\begin{pf}
The assertions follow from the definition of $c_1, c_2$.
\end{pf}
\subsection{The Jacobian matrix $\cal{J}$.}\label{jacob} Let $X_i=(x_{kl}^{(i)}),\
i=1,2$ denote the matrix of variables of size $n_2\times n_1$ (resp.
$n_3\times n_2$). Let
$\cal{M}_i=\{\text{ all } k_i+1
\text{ minors of } X_i\},\ i=1,2
$, $f_{rs}=(r,s)\text{ -th entry in }X_2X_{1},\ 1\le r\le n_3,\ 1\le s\le n_1.$
Let $\cal{J}$ be the Jacobian matrix of $V$ ($V$ considered as a subvariety of $Y,\
Y$ being as in \S 3 ). We shall index the rows of $\cal{J}$ by $\cal{M}_i,\ i=1,2,\
f_{rs}, \ 1\le r\le n_3,\ 1\le s\le n_1 $, and the columns by $\{x_{kl}^{(i)},\
1\le k\le n_{i+1},\ 1\le l\le n_{i},\ i=1,2\}$. Let $M \in \cal{M}_i,\ i=1,2$. The
entry in the $(M,x_{kl}^{(j)})\text{ -th position in } \cal{J}$ is non-zero if and
only if $j=i$ and $x_{kl}^{(i)}$ is an entry in $M$, in which case it is equal to
$\pm$ the $k_i$-minor of $M$ obtained by deleting the row and column through
$x_{kl}^{(i)}$. Also, for a variable $x_{ij}^{(1)}$, the only relevant functions are
$f_{mj},\ 1\le m \le n_3$. The $(f_{mj},x_{ij}^{(1)})$-th entry in $\cal{J}$ is
$x_{mi}^{(2)}$. Similarly, for a variable $x_{ij}^{(2)}$,  the only relevant
functions are $f_{im},\ 1\le m \le n_1$. The $(f_{im},x_{ij}^{(2)})$-th entry in
$\cal{J}$ is $x_{jm}^{(1)}$. From this it follows that  given $j,\ 1\le j\le n_1$,
the block in $\cal{J}$ with row indices given by $f_{rs}, \ 1\le r\le n_3$ (for a
fixed $s$) and column indices given by the variables in the
$j$-th column of $X_1 $ is the zero block if $s\ne j$, and for $s=j$, it is simply
$X_2$. Similarly, given $j,\ 1\le j\le n_3$,
the block in $\cal{J}$ with row indices given by $f_{sr}, \ 1\le r\le n_1$ (for a
fixed $s$) and column indices given by the variables in the
$j$-th row of $X_2 $ is the zero block if $s\ne j$, and for $s=j$, it is simply
$^tX_1$.  Given $x \in V$, let $\cal{J}_x$ denote $\cal{J}$ evaluated at
$x$.
\begin{lem}\label{rank}
Let $x\in O(t_1,t_2)$. Let $\cal{J}_0$ be the
submatrix of $\cal{J}_x$ consisting of the rows indexed by
$\{f_{rs},\ 1\le r\le n_3,\ 1\le s\le n_1\}$. Then rank$J_0=n_1t_2+n_3t_1-t_1t_2$.
\end{lem}

\begin{pf}
Let $x=(R_1,R_2)$. We may take $x$ to be the point of $O(t_1,t_2)$ given as follows:
the right hand bottom block of size $t_1$ in $R_1$ is Id$_{t_1}$ and the rest of the
entries in
$R_1$ are zero while the left hand top
block of size $t_2$ in $R_2$ is Id$_{t_2}$ and the rest of the entries in $R_2$ are
zero. Note that rank$R_1=t_1$, rank$R_2=t_2$, $R_2R_1=0$ (since $n_2\ge
t_1+t_2$). We shall denote by $M_1$ (resp. $M_2$) right hand bottom
block of size $t_1$ in $R_1$ (resp. the left hand top
block of size $t_2$ in $R_2$).

Let $ \cal{B}_i=\{ji,\ 1\le j\le t_2\},\ 1\le i \le n_1$, and $\cal{C}_i=\{ij,\ 1\le
j \le n_1\},\ t_2+1\le i\le n_3 $. We shall index the rows of $\cal{J}_0$ as
$\cal{B}_1, \cal{B}_2, \cdots , \cal{B}_{n_1},\ \cal{C}_{t_2+1}, \cal{C}_{t_2+2},
\cdots ,\cal{C}_{n_3} $. Let $J_1$
(resp. $J_2$) be the submatrix of $\cal{J}_0$ consisting of the rows of
$\cal{J}_0$ indexed by $\cal{B}_1,
\cal{B}_2, \cdots ,
\cal{B}_{n_1}$ (resp. $\cal{C}_{t_2+1}, \cal{C}_{t_2+2},
\cdots ,\cal{C}_{n_3} $).

We have (cf. \S \ref{jacob}), given $j,\ 1\le j\le n_1$, the block in
$J_1$ with row indices given by $\cal{B}_k$ and column indices given by the variables
in the
$j$-th column of $R_1 $ is the zero block if $k\ne j$, and for $k=j$, it is simply the
submatrix of
$R_2$ with row indices  given by $1,2, \cdots ,t_2$, and column indices given
by
$1,2,
\cdots ,n_2$ (and this block has rank $t_2$); similarly the block in $J_2$ with
row indices
$\{f_{rk},\ t_2+1\le r\le n_3\}$ (for a fixed $k$) and column indices given by the
variables in the
$j$-th column of $R_1,\ 1\le i\le n_1 $ is the zero block if $k\ne j$, and for $k=j$,
it is simply the submatrix of $R_2$ consisting of the last $(n_3-t_2)$ rows of
$R_2$ and hence is the zero block by our choice of $R_2$. Hence we obtain
$$\text{rank}\cal{J}_0=\text{rank}J_{1}+\text{rank}J_{2}.$$
From the discussion above, it follows easily that rank$J_1=n_1t_2$. and rank$J_2$
equals the rank of the submatrix of $J_2$ consisting of the columns indexed by
$x_{ij}^{(2)},\ 1\le i\le n_3,\ 1\le j\le n_2$, and rank$J_2=(n_3-t_2)t_1$.
Hence rank$J_0=n_1t_2+(n_3-t_2)t_1=n_1t_2+n_3t_1-t_1t_2$.
\end{pf}
\begin{prop}\label{4.3} Let $x=(A_1,A_2) \in V(k_1,k_2)$.
\begin{enumerate}
\item Let  rank $A_1=k_1$, rank $A_2<k_2$. Then $x$ is a smooth point of $V(k_1,k_2)\
\iff$ either $n_3=k_2$ or $n_2=k_1+k_{2}$.
\item Let  rank $A_2=k_2$, rank $A_1<k_1$. Then $x$ is a smooth point of $V(k_1,k_2)\
\iff$ either $n_1=k_1$ or $n_2=k_1+k_{2}$.
\item Let  rank $A_1=k_1-1$, rank $A_2=k_2-1$. Then $x$ is a singular point of
$V(k_1,k_2)$.
\end{enumerate}
\end{prop}
\begin{pf} Let $x=(A_1,A_2)$, and as above, let $\cal{J}_x$ denote $\cal{J}$ evaluated
at
$x$.

\ni (1) Let rank$A_2=t_2$. Then $x\in O(k_1,t_2)$. We may
take $x$ to be the point of $O(k_1,t_2)$ given as follows: the right hand bottom
block of size $k_1$ in $A_1$ is Id$_{k_1}$ and the rest of the entries in $A_1$ are
zero, while the left hand top
block of size $t_2$ in $A_2$ is Id$_{t_2}$ and the rest of the entries in $A_2$ are
zero. Note that rank$A_1=k_1$, rank$A_2=t_2$, $A_2A_1=0$ (since $n_2-k_1\ge
k_2>t_2$). We shall denote by $M_1$ (resp. $M_2$) right hand bottom
block of size $k_1$ in $A_1$ (resp. the left hand top
block of size $t_2$ in $A_2$). Let
$\cal{B}_1$ denote the set of all variables in $A_1$ not appearing in any of the
last $k_1$ rows and last $k_1$ columns of $A_1$. Let
$\t\in \cal{B}_1 $, and let $M_\t$ be the $(k_1+1)$-minor of $A_1$ obtained
from $M_1$ by adding the row and column of $A_1$ through $\t$. Then the
$(M_\t,\t)$-th entry in $\cal{J}_x$ is equal to $1$ ; also for $\sigma \in \cal{B}_1,
\ \sigma \ne \t$, the $(M_\t,\sigma)$-th entry in $\cal{J}_x$ is $0$. Further, for
$\t\not\in \cal{B}_1 $, the $(M,\t)$-th entry in $\cal{J}_x$ is zero, for all $M\in
\cal{M}_1$ (by our choice of $A_1$). Also, for any variable $\phi$ in $X_2$,
the $(M,\phi)$-th entry in $\cal{J}_x$ is zero, for all $M\in
\cal{M}_2$ (since rank$A_2<k_2$). Let
$J_1$
 be the submatrix of $\cal{J}_x$ consisting of the columns of
$\cal{J}_x$ indexed by the variables which are the entries of the first $(n_1-k_1)$
columns of
$X_1$ and $J_2$ the submatrix of $\cal{J}_x$ consisting of the remaining columns of
$\cal{J}_x$. Then clearly rank$\cal{J}_x=$ rank$J_1+$ rank$J_2$ (by our choice of
$A_1$).

\ni {\bf Computation of rank$J_1$.} We shall index the columns of $J_1$ by starting
with the variables in the first column of $X_1$, followed by the variables in the
second column, $\ldots$, followed by the variables in the $(n_1-k_1)$-th column of
$X_1$. Let $J_{1i}$ be the submatrix of $J_1$ consisting the columns of $J_1$ indexed
by the variables in the $i$-th column of $X_1,\ 1\le i\le n_1-k_1$.
Then by our choice of $A_1,A_2$,  the non-zero entries in $J_{1i}$ occur at the
$(M_\t,\t)$-th place,
$\t \in \{x_{1i}^{(1)}, x_{2i}^{(1)},
\cdots ,x_{n_2-k_1\,i}^{(1)}\}$, $(f_{ji},x_{ji}^{(1)})$-th place, $1\le j \le t_2$
(and these entries are equal to $1$; note that the $(f_{ji},x_{ji}^{(1)})$-th is
$x_{jj}^{(2)}$).  But now the fact that $ t_2<n_2-k_1$ implies that the first
$(n_2-k_1)$ columns of
$J_{1i}$ are the only non-zero columns of $J_{1i}$ and these are clearly linearly
independant. Hence we obtain rank$J_{1i}=n_2-k_1,\ 1\le i\le n_1-k_1$, and hence
rank$J_1=\sum_{i=1}^{n_1-k_1}\ \text{rank}J_{1i}=(n_1-k_1)(n_2-k_1)=c_1$.

\ni {\bf Computation of rank$J_2$.} By our choice of $A_1$, we have, rank$J_2$
equals the rank of the submatrix $J_3$ of $J_2$ consisting of the rows indexed by
$\{f_{rs},\ 1\le r\le n_3,\ n_1-k_1+1\le s\le n_1\}$. Taking $R_1$ of
Lemma \ref{rank}  as the submatrix of $A_1$ consisting of the last $k_1$ columns of
$A_1$ and
$R_2$ as $A_2$, we have, rank$J_2=k_1t_2+n_3k_1-k_1t_2=n_3k_1$ (note that in Lemma
\ref{rank}, $n_1=k_1$ since $R_1$ has size $n_2\times k_1$, and $t_1=k_1$, since
rank$R_1=k_1$).

Hence we obtain
$$\text{rank }\cal{J}_x=c_1+n_3k_1.$$

If $n_3=k_2$ or $n_2=k_1+k_{2}$, then
$n_3k_1=c_2+ k_1k_2$ (cf. Lemma
\ref{4.1}, (2)). Hence $\text{rank
}\cal{J}_x=c_1+c_2+k_1k_2=\text{ codim}_Y V(k_1,k_2)$, and hence $x$ is a smooth
point of $V(k_1,k_2)$.

Let $n_3>k_2$ and $n_2>k_1+k_{2}$. We have, $\text{rank
}\cal{J}_x=c_1+n_3k_1=c_1+c_2+k_1k_2-(n_3-k_2)(n_2-k_1-k_2)$ (cf. Lemma \ref{4.1},
(2)). This together with the hypothesis that $n_3>k_2,\  n_2>k_1+k_{2}$ implies that
$\text{rank
}\cal{J}_x<c_1+c_2+k_1k_2\, (=\text{ codim}_Y V(k_1,k_2))$. Hence $x$ is a singular
point of  $V(k_1,k_2)$.

\ni (2) The proof of (2) is similar. Let rank$A_1=t_1$. Then $x\in O(t_1,k_2)$. We may
take $x$ to be the point of $O(t_1,k_2)$ given as follows: the right hand bottom
block of size $k_2$ in $A_2$ is Id$_{k_2}$ and the rest of the entries in $A_2$ are
zero while the left hand top
block of size $t_1$ in $A_1$ is Id$_{t_1}$ and the rest of the entries in $A_1$ are
zero. Note that rank$A_1=t_1$, rank$A_2=k_2$, $A_2A_1=0$ (since $n_2-k_2\ge
k_1>t_1$). We shall denote by $M_1$ (resp. $M_2$) left hand top
block of size $t_1$ in $A_1$ (resp. the right hand bottom
block of size $k_2$ in $A_2$). Let
$\cal{B}_2$ denote the set of all variables in $A_2$ not appearing in any of the
last $k_2$ rows and last $k_2$ columns of $A_2$. Let
$\t\in
\cal{B}_2 $, and let $M_\t$ be the $(k_2+1)$-minor of $A_2$ obtained
from $M_2$ by adding the row and column of $A_2$ through $\t$. Then the
$(M_\t,\t)$-th entry in $\cal{J}_x$ is equal to $1$ ; also for $\sigma \in \cal{B}_2,
\ \sigma \ne \t$, the $(M_\t,\sigma)$-th entry in $\cal{J}_x$ is $0$. Let $J_1$
 be the submatrix of $\cal{J}_x$ consisting of the columns of
$\cal{J}_x$ indexed by the variables which are the entries of the first $(n_3-k_2)$
rows of
$X_2$ and $J_2$ the submatrix of $\cal{J}_x$ consisting of the remaining columns of
$\cal{J}_x$. Then clearly rank$\cal{J}_x=$ rank$J_1+$ rank$J_2$ (by our choice of
$A_2$).

\ni {\bf Computation of rank$J_1$.} We shall index the columns of $J_1$ by starting
with the variables in the first row of $X_2$, followed by the variables in the
second row, $\ldots$, followed by the variables in the $(n_3-k_2)$-th row of
$X_2$. Let $J_{1i}$ be the submatrix of $J_1$ consisting the columns of $J_1$ indexed
by the variables in the $i$-th row of $X_2,\ 1\le i\le n_3-k_2$.
Then by our choice of $A_1,A_2$, the non-zero entries in $J_{1i}$ occur at the
$(M_\t,\t)$-th place,
$\t \in \{x_{i1}^{(2)}, x_{i2}^{(2)},
\cdots ,x_{i\,n_3-k_2}^{(2)}\}$, $(f_{ij},x_{ij}^{(2)})$-th place, $1\le j \le t_1$
(and these entries are equal to $1$; note that the $(f_{ij},x_{ij}^{(2)})$-th
entry is $x_{jj}^{(1)}$). But now the fact that $ t_1<n_2-k_2$ implies that the
first $(n_2-k_2)$ columns of
$J_{1i}$ are the only non-zero columns of $J_{1i}$ and these are clearly linearly
independant. Hence we obtain rank$J_{1i}=n_2-k_2,\ 1\le i\le n_3-k_2$, and hence
rank$J_1=\sum_{i=1}^{n_3-k_2}\ \text{rank}J_{1i}=(n_2-k_2)(n_3-k_2)=c_2$.

\ni {\bf Computation of rank$J_2$.} By our choice of $A_2$, we have, rank$J_2$
equals the rank of the submatrix $J_3$ of $J_2$ consisting of the rows indexed by
$\{f_{rs},\ n_3-k_2+1\le r\le n_3,\ 1\le s\le n_1\}$. Taking $R_2$ of
Lemma \ref{rank}  as the submatrix of $A_2$ consisting of the last $k_2$ rows of $A_2$
and
$R_1$ as $A_1$, we have, rank$J_2=n_1k_2+k_2t_1-t_1k_2=n_1k_2$ (note that in Lemma
\ref{rank}, $n_3=k_2$ since $R_2$ has size $k_2\times n_2$, and $t_2=k_2$, since
rank$R_2=k_2$).

Hence we obtain
$$\text{rank }\cal{J}_x=c_2+n_1k_2.$$

If $n_1=k_1$ or $n_2=k_1+k_{2}$, then
$n_1k_2=c_1+k_1k_{2}$ (cf. Lemma \ref{4.1}, (1)). Hence
$\text{rank }\cal{J}_x=c_1+c_2+k_1k_2=\text{ codim}_Y V(k_1,k_2)$ , and $x$ is a
smooth point of $V(k_1,k_2)$.

If $n_1>k_1$ and $n_2>k_1+k_{2}$, then $\text{rank }\cal{J}_x<\text{codim}_Y
V(k_1,k_2)$ (in view of Lemma \ref{4.1},(1)). Hence $x$ is a singular point of  $V(k_1,k_2)$.

\ni (3) Let  rank $A_1=k_1-1$, rank $A_2=k_2-1$. Let us denote $t_1=k_1-1,\
t_2=k_2-1$. Then $x\in O(t_1,t_2)$. We have, the block in
$\cal{J}_x$  corresponding to  the rows indexed by $\cal{M}_1$ and columns indexed by
the set of all variables of $A_1$ is the zero block (since rank $A_1<k_1$), and the block in $\cal{J}_x$ corresponding to the rows indexed by $\cal{M}_2$ and columns
indexed by the set of all variables of $A_2$ is the zero block (since
rank$A_2<k_2$). Hence we obtain that $\text{rank}\cal{J}_x$ equals the rank of  the
submatrix $\cal{J}_0$ in $\cal{J}_x$ (consisting of the rows indexed by
$\{f_{rs},\ 1\le r\le n_3,\ 1\le s\le n_1\}$). Now Lemma \ref{rank} (with
$R_1=A_1,\ R_2=A_2$) implies that rank
$\cal{J}_x=n_1t_2+n_3t_1-t_1t_2$. This implies codim$_Y
V(k_1,k_2)-\text{ rk}
\cal{J}_x=c_1+c_2+k_1k_2-(n_1t_2+n_3t_1-t_1t_2)$

\ni $=c_1+c_2+k_1k_2-[ c_1+k_1k_2-
(n_1-k_1)(n_2-k_1-k_2)+c_2+k_1k_2-(n_3-k_2)(n_2-k_1-k_2)]+n_1-n_3+(k_1-1)(k_2-1)$
 (cf. Lemma \ref{4.1}) $=(n_1+n_3-k_1-k_2)(n_2-k_1-k_2+1)+1>0$ (note that
$n_1+n_3-k_1-k_2=(n_1-k_1)+(n_3-k_2)\ge 0$. From this it follows that $x$ is a
singular point of  $V(k_1,k_2)$.
\end{pf}
\begin{thm}\label{4.4} Let Sing$V(k_1,k_2)$ denote the singular locus of $V(k_1,k_2)$.
We have
$$\text{Sing}V(k_1,k_2)=
\begin{cases}
V(k_1-1,k_2-1), \text{ if either } & k_1+k_2=n_2 \text{ or } k_1=n_1,\ k_2=n_3\\
V(k_1-1,k_2),& k_1+k_2<n_2 ,\ k_1<n_1,\,k_2=n_3\\
V(k_1,k_2-1),& k_1+k_2<n_2,\ k_2<n_3,\,k_1=n_1\\
V(k_1-1,k_2)\cup V(k_1,k_2-1),& k_1+k_2<n_2 ,\ k_1<n_1,\,k_2<n_3.
\end{cases}$$
\end{thm}
\begin{pf}
 Let $x=(A_1,A_2)\in V(k_1,k_2)$. If rk
$A_i=k_i,\ i=1,2$, then $x\in Z^{\circ}(k_1,k_2)$ (= the $G_{\bold{n}}$-orbit
through $x$), and hence $x$ is a smooth point of $V(k_1,k_2)$ (note that $V(k_1,k_2)={{\overline
{Z^0(k_1,k_2)}}}$ ). Thus we obtain (using Proposition \ref{4.3}, (3))
$$V(k_1-1,k_2-1)\subseteq \text{ Sing }V(k_1,k_2)\subseteq V(k_1-1,k_2)\cup
V(k_1,k_2-1).$$
The result follows from this (in view of Proposition \ref{4.3}).
\end{pf}

\begin{lem}\label{5.1}
$V(k_1,k_2)$ being an irreducible component of $V$, we have, either $k_1+k_2=n_2
\text{ or } k_1=n_1,\ k_2=n_3$.
\end{lem}
\begin{pf}
 Note that $(k_1,k_2)$ is a maximal element of
$A_{\bold{n}}$ (for the partial order on $A_{\bold{n}}$ defined in \S \ref{3.1}). We
divide the proof into the following two cases.

\ni {\bf Case 1.} $n_2\ge n_1+n_3$.

\ni In this case, there is a unique maximal choice for $k_1,k_2$, namely $k_1=n_1,\
k_2=n_3$.

\ni {\bf Case 2.} $n_2< n_1+n_3$.

\ni If $n_2\ge n_1$, then $(k_1,k_2)=(n_1-i, n_2-n_1+i),\ i=0,1,\cdots , n_1$
are all the possible maximal choices for $(k_1,k_2)$, and for all such choices,
we have, $k_1+k_2=n_2$.

\ni Similarly, if $n_2\ge n_3$, then $(k_1,k_2)=( n_2-n_3+i,n_3-i),\ i=0,1,\cdots ,
n_3$ are all the possible maximal choices for $(k_1,k_2)$, and for all such choices,
we have, $k_1+k_2=n_2$.

\ni If $n_2< n_1,\ n_2< n_3$, then $(k_1,k_2)=(i, n_2-i),\ i=0,1,\cdots ,
n_2$ are all the possible maximal choices for $(k_1,k_2)$, and for all such choices,
we have, $k_1+k_2=n_2$.
\end{pf}
\begin{rem}\label{5.2'}
From the proof of Lemma \ref{5.1}, we have, if $n_2\ge n_1+n_3$, then $k_1=n_1,\
k_2=n_3$; if $n_2< n_1+n_3$, then $k_1+k_2=n_2$.
\end{rem}
\begin{thm}\label{5.2}
$V(k_1,k_2)$ being an irreducible component of $V$, we have,
$\text{Sing}V(k_1,k_2)=V(k_1-1,k_2-1)$.
\end{thm}
\begin{pf}
We have (cf. Lemma \ref{5.1}), either $k_1+k_2=n_2 \text{ or } k_1=n_1,\ k_2=n_3$ from
which the result follows (in view of Theorem \ref{4.4}).
\end{pf}


%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{One Application.} We preserve the notations of sections \S 3,4. Let
$V(k_1,k_2)$ be an irreducible component of $V$, and let $w\in W^Q $ be the element
such that $V(k_1,k_2)\cong Y_Q(w)$. In this section, using Theorem \ref{5.2} we
deduce results on Sing$X(w^{\text{max}}),\ w^{\text{max}}$ being the maximal
representative in $W$ of the coset $wW_Q$ (note that $X(w^{\text{max}})$ is the
pull back of $X(w)$ under the canonical projection $G/B\rightarrow G/Q$).


%%%%%%%%%%%%%%%%%%%%%%%%
%%% conjecture

\subsection{A conjecture on the irreducible components of a Schubert variety in
$SL(n)/B$}\label{9} Let $G=SL(n)$. We recall (cf. \cite{l-s}) a
conjecture on the irreducible components of the singular locus of a  Schubert
variety.

For $\t\in W\,(=S_n)$, let $P_\t$ (resp. $Q_\t$) be the maximal element of the set of parabolic
subgroups which leave $\overline{B\t B}$ ($\subset G$) stable under multiplication on the left
(resp. right).


\begin{defn}
Given parabolic subgroups $P$, $Q$, we say that $\overline{B\t B}$ is $P$-$Q$ stable if $P\subset
P_\t$ and $Q\subset Q_\t$.
\end{defn}

\noindent {\bf The set} $F_{\eta}$.


Let $\eta=(a_1\dots a_n)\in\cal{S}_n$. Let $E_{\eta}$ be the set of all $\t_1\le\eta$ such that
either $1)$ or $2)$ below holds.

\noindent $1)$ There exist $i,j,k,m$, $1\le i<j<k<m\le n$, such that

$(a)$ $a_k<a_m<a_i<a_j$

$(b)$ if $\t_1=(b_1\dots b_n)$, then there exist  $i',j',k',m'$, $1\le i'<j'<k'<m'\le n$ such
that $b_{i'}=a_k$, $b_{j'}=a_i$, $b_{k'}=a_m$, $b_{m'}=a_j$

$(c)$ if $\t$ (resp. $\te_1$) is the element obtained from $\eta$ (resp. $\t_1$) by replacing
$a_i,a_j,a_k,a_m$ respectively by $a_k,a_i,a_m,a_j$ (resp. $b_{i'},b_{j'},b_{k'},b_{m'}$
respectively by $b_{j'},b_{m'},b_{i'},b_{k'}$), then $\t_1\ge\t$ and $\te_1\le\eta$.

\noindent $2)$ There exist $i,j,k,m$, $1\le i<j<k<m\le n$, such that

$(a)$ $a_m<a_j<a_k<a_i$

$(b)$ if $\t_1=(b_1\dots b_n)$, then there exist  $i',j',k',m'$, $1\le i'<j'<k'<m'\le n$ such
that $b_{i'}=a_j$, $b_{j'}=a_m$, $b_{k'}=a_i$, $b_{m'}=a_k$

$(c)$ if $\t$ (resp. $\te_1$) is the element obtained from $\eta$ (resp. $\t_1$) by replacing
$a_i,a_j,a_k,a_m$ respectively by $a_j,a_m,a_i,a_k$ (resp. $b_{i'},b_{j'},b_{k'},b_{m'}$
respectively by $b_{k'},b_{i'},b_{m'},b_{j'}$), then $\t_1\ge\t$ and $\te_1\le\eta$.

Let $F_\eta=\{\t\in E_\eta\mid\overline{B\t B}\text{ is }P_\eta\text{-}Q_\eta\text{ stable}\}$.
\begin{conj}\label{conjecture}
Let $X(\eta)= \overline{B\eta B}$ (mod $B$) ($\subset G/B$). The singular locus of
$X(\eta)$ is equal to $\cup_{\l}X(\l)$, where
$\l$ runs over the maximal (under the Bruhat order) elements of $F_{\eta}$.
\end{conj}



\subsection{}
Let $\eta=(a_1\dots a_n)\in\cal{S}_n$. Let $\text{Sing\,}X(\eta)\ne \emptyset$. Let $(a,b,c,d)$
be four distinct entries in $\{1,\dots,n\}$ such that $a<b<c<d$. An occurence in
$\eta$ of the form $d,b,c,a$, where $d=a_i$, $b=a_j$, $c=a_k$, $a=a_m$, $i<j<k<m$, will be
referred to as a {\em Type I bad pattern in } $\eta$. An occurance in $\eta$ of the
form
$(c,d,a,b)$, where
$c=a_i$, $d=a_j$, $a=a_k$, $b=a_m$, $i<j<k<m$, will be referred to as a {\em Type II
bad pattern in} $\eta$. Recall (cf. \cite{l-s}) that $\text{Sing\,}X(\eta)\ne
\emptyset$ if and only if there is a bad pattern of Type I or II in $\eta$.

%%%%%%%%%%%%%%%%%%%%%%%%
\subsection{ Application to Schubert varieties.} Let $V(k_1,k_2)$ be an irreducible
component of
$V$. Let
$a_1=n_1,\,a_2=n_1+n_2,\,a_3=n_1+n_2+n_3\,(=n)$, $G=SL(n)$, and $Q=P_{a_1}\cap P_{a_2} $. We have (cf.
Theorem \ref{2.4}) $V(k_1,k_2)\simeq Y_Q(w),\ V(k_1-1,k_2-1)\simeq Y_Q(\te)$, where
$w,
\te
\in W^Q$ are given by
$$w^{(a_1)}=(\,\underbrace{\ldots a_{1}}_
{\mbox{\small $a_{1}-k_1$}}
\ \underbrace{. \ldots a_{2}}_
{\mbox{\small $k_1$}}),\
w^{(a_2)}=(\,\underbrace{1\ldots a_{1}}_
{\mbox{\small $a_{1}$}}
\ \underbrace{. \ldots a_{2}}_
{\mbox{\small $n_2-k_2$}}\  \underbrace{. \ldots a_{3}}_
{\mbox{\small $k_2$}}), $$
$$\te^{(a_1)}=(\,\underbrace{\ldots a_{1}}_
{\mbox{\small $a_{1}-k_1+1$}}
\ \underbrace{. \ldots a_{2}}_
{\mbox{\small $k_1-1$}}),\
\te^{(a_2)}=(\,\underbrace{1\ldots a_{1}}_
{\mbox{\small $a_{1}$}}
\ \underbrace{. \ldots a_{2}}_
{\mbox{\small $n_2-k_2+1$}}\  \underbrace{. \ldots a_{3}}_
{\mbox{\small $k_2-1$}}). $$
\subsection{}\label{cases} We have the following two cases (cf. proof of Lemma
\ref{5.1}):

\ni {\bf Case1.} $n_2\ge n_1+n_3$.

\ni In this case, we have $k_1=n_1,\
k_2=n_3$ (cf. Remark \ref{5.2'}). Hence
$$w^{(a_1)}=([x,a_2]),\ w^{(a_2)}=([1,a_1],[x',a_2],[y,a_3]),$$
$$\te ^{(a_1)}=([n_1],[x+1,a_2]),\ \te ^{(a_2)}=([1,a_1],[x'-1,a_2],[y+1,a_3]),$$
where
$x=a_2+1-k_1\,(=n_2+1),\ x'=a_2+1-(n_2-k_2)\,(=n_1+n_3+1),\ y=a_3+1-k_2\,(=a_2+1)$.

\ni {\bf Case 2.} $n_2< n_1+n_3$.

\ni In this case, we have, $k_1+k_2=n_2$ (cf. Remark \ref{5.2'}). Hence
$$w^{(a_1)}=([p,a_1],[x,a_2]),\ w^{(a_2)}=([1,a_1],[x,a_2],[y,a_3]),$$
$$\te ^{(a_1)}=([p-1,a_1],[x+1,a_2]),\ \te
^{(a_2)}=([1,a_1],[x-1,a_2],[y+1,a_3]),$$ where
$p=a_1+1-(n_1-k_1),\ x=a_2+1-k_1\,(=n_1+k_2+1),\  y=a_3+1-k_2$.

\begin{rem}\label{7.2}
Note that $\te$ is the unique maximal element $\t$ in $W^Q$ such that
$\t \le w$ and $\t ^{(a_i)}(a_i-k_i+1)\le a_i,\ i=1,2$ (here, $\t ^{(a_i)}(t)$ denotes the entry at the $t$-th
place in $\t ^{(a_i)}$).
\end{rem}


\subsection{} Let $Q$ be as
above. As above, we shall denote an element $\t$ in $W^Q$ as $\t=(\t ^{(a_1)},\t
^{(a_2)})$. Let $\t ^{\text{max}}$ denote the maximal representative in $W$ of the
coset $\t W_Q$. Note that $X(\t ^{\text{max}})$) is the inverse image of $X_Q(\te )$)
under the canonical projection $G/B\rightarrow G/Q$; also, $\t^{\text{max}}$ as a
permutation is given by writing the entries in $\t ^{(a_1)}$ in descending order,
followed by the entries in $\t ^{(a_2)} \setminus \t ^{(a_1)}$ in descending order,
followed by the entries in $\{1,\cdots ,n\}\setminus \t ^{(a_2)} $ in descending
order. In the sequel, we shall refer to these as the I,II,III blocks respectively of
$\t ^{\text{max}}$. Also, in the sequel, for an interval $[a,b],\ [a,b]_{-}$ shall
denete the entries of $[a,b]$ written in descending order. Thus, in Case 1 of
\ref{cases}, we have (with notations as in \ref{cases}),

\ni I block in $w^{\text{max}}=[n_2+1,a_2]_{-}$, I block in $\te ^{\text{max}}=([n_2+2,a_2]_{-},[n_1])$,

\ni II block in $w^{\text{max}}=([y,a_3]_{-},\,[x',n_2]_{-},\,[1,a_1]_{-}),
$ II block in $\te
^{\text{max}}=([y+1,a_3]_{-},\,[x'-1,n_2+1]_{-},\,[1,a_1-1]_{-}),$

\ni III block in $w^{\text{max}}=[a_1+1,x'-1]_{-},$ III block in $\te
^{\text{max}}=([a_2+1],[a_1+1,x'-2]_{-})$.

\ni In Case 2 of
\ref{cases}, we have (with notations as in \ref{cases}),

\ni I block in $w^{\text{max}}=([x,a_2]_{-},[p,a_1]_{-}),$ I block in $\te
^{\text{max}}=([x+1,a_2]_{-},[p-1,a_1]_{-}),$

\ni II block in $w^{\text{max}}=([y,a_3]_{-},[1,p-1]_{-}),$ II block in $\te
^{\text{max}}=([y+1,a_3]_{-},\,[x-1,x]_{-},\,[1,p-2]_{-}),$

\ni III block in $w^{\text{max}}=([a_2+1,y-1]_{-},[a_1+1,x-1]_{-}),$ III block in $\te
^{\text{max}}=([a_2+1,y]_{-},[a_1+1,x-2]_{-})$.

\ni For $\t \in W$, we shall denote the associated Schubert variety by $X(\t)$.

\begin{thm}
Sing $X(w^{\text{max}})=X(\te^{\text{max}})$
\end{thm}
\begin{pf}
The result follows from Theorems \ref{2.4}, \ref{5.2}.
\end{pf}
We now show that the Conjecture 6.2  holds for $X:=X(w^{\text{max}})$.
\begin{prop}\label{7.5}
A bad pattern in $w^{\text{max}}$ has to be of Type II.
\end{prop}
\begin{pf}
If possible, assume that there is a Type I bad pattern $d,b,c,a$ in $w^{\text{max}}$,
where $a<b<c<d$. We have,
$d$ belongs to the I or II block in $w^{\text{max}}$.

\ni (1) Let $d$ belong to the I block in $w^{\text{max}}$. This implies that $b \in
$ I or II block.

Let $b$ belong to the I block in which case we have $c$ does not belong to the I
block. In Case 1 of
\ref{cases}, we have,
$[b,d]\subset$ I block in $w^{\text{max}}$, and  this contradicts the facts that
$b<c<d$ and $c$ does not belong to the I block. Hence our
assumption is wrong and the result follows. In Case 2 of
\ref{cases}, we have, $d\ge x,\ p\le b\le a_1$. Hence we obtain $a\le p$, and hence
$a,c$ should belong to the II block in $w^{\text{max}}$. This in turn implies that $c
\in [y,a_3]$, which is not possible since $c<d$, and $d<y$. The result follows from
this.

Let now $b$ belong to the II block in $w^{\text{max}}$ in which case we have $c,a$ should
belong to the III block. In Case 1 of
\ref{cases}, we have, $[a,c]$ is contained in the III block which is not possible,
since $a<b<c$ and $b$ does not belong to the III
block. In Case 2 of
\ref{cases}, we have, $b\in [1,p-1]$, since $b<d<y$. But now, $a\ge a_1+1\, (>p-1)$
(since
$a$ belongs to the III block), which contradicts the fact that $b>a$.

\ni (2) Let $d$ belong to the II block in $w^{\text{max}}$. This implies that $b$ belongs to the II block,
and $c,a$ belong to the III block. In Case 1 of
\ref{cases}, we have, $[a,c]\subset$ III block, which is not possible, since $a<b<c$ and $b$ does not belong to
the III block. In Case 2 of
\ref{cases}, we have, $b\in [1,p-1]$ necessarily (since $b<c<y$) in which case, we
obtain $a>b$ (since $a\ge a_1+1>p-1$), which is not possible.
\end{pf}
\begin{prop}\label{6.6}
There do exist Type II bad pattern in $w^{\text{max}}$. Further for any Type II bad
pattern $c,d,a,b$ in $w^{\text{max}}$, where $a<b<c<d$, we have, $c\in [x,a_2],\
d\in [y,a_3],\ a\in [1,u],\ b\in [a_1+1,v]$, where $x=$ or $>n_2+1$ according as
$n_1=$ or $>k_1$, $y=$ or $>a_2+1$ according as
$n_3=$ or $>k_2$, $u=a_1$ (respectively $p-1$) in Case 1 (respectively Case 2) of
\ref{cases}, $v=(x'-1)$ (respectively $x-1$) in Case 1 (respectively Case 2) of
\ref{cases}.
\end{prop}
\begin{pf} We have a Type II pattern in $w^{\text{max}}$, namely,
$c=x,\ d=y,\ a=u,\ b=v$ in $w^{\text{max}}$ (here,
$x,y,u,v$ are as in Proposition \ref{6.6}).

Let now $c,d,a,b $ be a Type II pattern in $w^{\text{max}}$.
We have $c$ belongs to the I block, $d,a$ belong to the II block, and $b$ belongs to
the III block (necessarily). We distinguish the following two cases:

\ni {\bf Case 1.} $n_2\ge n_1+n_3$.

\ni In this case we have $n_1=k_1$, and clearly $c\in
[n_2+1,a_2]$. This implies $d\in [y,a_3],\ a\in [1,a_1],\ b\in [a_1+1,x'-1]$.

\ni {\bf Case 2.} $n_2< n_1+n_3$.

\ni Again clearly $c\in
[x,a_2]$. This implies $d\in [y,a_3],\ a\in [1,p-1],\ b\in [a_1+1,x-1]$.
\end{pf}
\begin{cor}
With notations as in Proposition \ref{6.6}, we have, $w^{\text{max}}$ is
non-vexillary.
\end{cor}
\begin{pf}
This follows (in view of Proposition \ref{6.6}) from the definition of a vexillary
permutation; recall (cf.
\cite{l}) that a permutation $(a_1,\cdots ,a_n)$ is vexillary if it avoids the
pattern
$c,d,a,b$, where $a<b<c<d$.
\end{pf}
\begin{lem}\label{6.7}
\begin{enumerate}
\item $Q_{w^{\text{max}}}=P_{w^{\text{max}}}=Q$.
\item ${\overline {B\te ^{\text{max}}}}B$ is $Q$-$Q$ is stable , i.e., stable for
multiplication on the left and right by $Q$.
\end{enumerate}
\end{lem}
\begin{pf}
The assertions follow from the definition of $w^{\text{max}}$ and $\te ^{\text{max}}$
\end{pf}
\begin{thm}\label{7.3}
The Conjecture 6.2 \ref{conjecture} holds for $w^{\text{max}}$.
\end{thm}
\begin{pf}
We first observe that $\te ^{\text{max}}\in F_{w^{\text{max}}}$; for, corresponding
to the Type II pattern $c=x,\ d=y,\ a=u,\ b=v$ in $w^{\text{max}}$ (here,
$x,y,u,v$ are as in Proposition \ref{6.6}), we have the bad pattern $a,c,b,d$ in
$\te ^{\text{max}}$. Let us denote $\te ^{\text{max}}$ by $\phi$. Let $w_1$ (resp. $\t$) be the
element of
$S_n$ obtained from
$\phi$ (resp. $w^{\text{max}}$) by replacing $a,c,b,d$ (resp. $c,d,a,b$) respectively by $c,d,a,b$
(resp.
$a,c,b,d$). Then clearly $\t\le\phi$, and $w_1\le w$. Further,
$\overline{B\te^{\text{max}}B}$ is
$Q$-$Q$ stable (cf. Lemma \ref{6.7}). Thus $\te^{\text{max}}\in F_{w^{\text{max}}}$.

Let now $\t_1\in F_{w^{\text{max}}}$.

We have an occurrance in $\t_1$ of the form $(a,c,b,d)$,
$a<b<c<d$, corresponding to the occurrance $(c,d,a,b)$ in $w^{\text{max}}$ (cf.
Proposition \ref{7.5}). The fact that
$\t_1\in F_{w^{\text{max}}}$ in particular implies that
$\t_1 \in W_Q^{\text{max}} $, and this in turn implies that
$a$ belongs to the I block, $c,b$ belong to the II block, and $d$ belongs to the III
block in
$\t_1 $, i.e., $a\in \t_1 ^{(a_1)},\ b,c \in  \t_1 ^{(a_2)}$. Let $w_1$ be the
element of $S_n$ obtained from
$\t_1$ by replacing
$a,c,b,d$ respectively by
$c,d,a,b$. We have $w_1\le w^{\text{max}}$ (by definition of $F_{w^{\text{max}}}$).

Let $i$ (respectively $j$) be such that $w_1 ^{(a_1)}(i)\le a_1,\ w_1
^{(a_1)}(i+1)> a_1,\ w_1 ^{(a_1)}(j)=c$. Note that $i\ge a_1-k_1$ (since $w_1\le
w^{\text{max}}$ and $w^{(a_1)}(a_1-k_1)\le a_1$), $j\ge i+1$ (since
$c\ge x\,(=a_2+1-k_1)>n_1\,(= a_1)$). Now, $\t_1 ^{(a_1)}$ is obtained from $w_1
^{(a_1)}$ by replacing $c$ by
$a$, where note that $a\le a_1$ (cf. Proposition \ref{6.6}). From this we obtain
$\t_1 ^{(a_1)}(i+1)=w_1 ^{(a_1)}(i)\le a_1$ and hence
\begin{equation*}
\t_1 ^{(a_1)}(a_1-k_1+1)\le a_1\tag{1}
\end{equation*}
(note that $a_1-k_1+1\le i+1$ ). Let $k$ (respectively $l$) be such that $w_1
^{(a_2)}(k)\le a_2,\ w_1 ^{(a_2)}(k+1)> a_2,\ w_1 ^{(a_2)}(l)=d$. Note that $k\ge
a_2-k_2$ (since
$w_1\le w^{\text{max}}$ and $w^{(a_2)}(a_2-k_2)\le a_2$), $l\ge k+1$ (since $d\ge y>
a_2$ (cf. Proposition
\ref{6.6})). Now, $\t_1 ^{(a_2)}$ is obtained from $w_1 ^{(a_2)}$ by replacing $d$ by
$b$, where note that $b\le a_2$. From this we obtain $\t_1 ^{(a_2)}(k+1)=w_1
^{(a_2)}(k)\le a_2$ and hence
\begin{equation*}
\t_1 ^{(a_2)}(a_2-k_2+1)\le a_2\tag{2}
\end{equation*}
(note that $a_2-k_2+1\le k+1$ ). Now (1) and (2) together with Remark \ref{7.2} imply that $\t_1\le
\te^{\text{max}}$. The required result follows from this.
\end{pf}

\ni {\bf Example:} Let $\bold{n}=(2,3,2)$. We have $n=7,\ Q=P_2\cap P_5$. Further,
$V(1,2),\ V(2,1)$ are the two  irreducible components of $V$. Let $w_1,w_2$ be the
elements of $W^Q$ such that $V(1,2)\cong Y_Q(w_1),\ V(2,1)\cong Y_Q(w_2)$. We have,
$$w_1^{(2)}=(2,5),\ w_1^{(5)}=(1,2\,5\,67),$$
$$w_2^{(2)}=(4,5),\ w_2^{(5)}=(1,2\,4,5\,7).$$
 Let $\te_1,\te_2$ be the
elements of $W^Q$ such that $V(0,1)\cong Y_Q(\te_1),\ V(1,0)\cong Y_Q(\te_2)$. We
have,
$$\te_1^{(2)}=(1,2),\ \te_1^{(5)}=(1,2\,4,5\,7),$$
$$\te_2^{(2)}=(2,5),\ \te_2^{(5)}=(1,2\,3,4,5).$$
Further, we have,
$$w_1^{\text{max}}=(5276143),\ \te_1^{\text{max}}=(2175463),$$
$$w_2^{\text{max}}=(5472163),\ \te_2^{\text{max}}=(5243176).$$
Note that $w_1$ (resp. $w_2$) is non-vexillary with $5,6,1,4$ (resp. $4,7,2,3$) as a
Type II occurance. Note also that the occurance $1,5,4,6$ in $\te_1$ (resp.
$2,4,3,7$ in $\te_2$) corresponds to the occurance $5,6,1,4$ in $w_1$ (resp.
$4,7,2,3$ in
$w_1$).

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\end{thebibliography}


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