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\def\v{\vee}
\def\w{\wedge}
\def\mn{\underline{m}}
\def\ns{\underline{n}}
\def\st{\underline{s}}
\def\lt{\triangleleft}
\def\lte{\underline{\lt}}
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\def\L{\cal{L}}
\def\A{\cal{A}}
\def\C{\cal{C}}
\def\e{\varepsilon}
\def\a{\alpha}
\def\t{\tau}
\def\b{\beta}
\def\f{\phi}
\def\l{\lambda}
\def\te{\theta}
\def\s{\sigma}
\def\o{\omega}
\def\d{\delta}
\def\g{\gamma}
\def\h{\chi}
\def\m{\mu}
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\def\ui{\underline{i}}
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\begin{document}
\title[Schubert varieties, toric varieties, and  determinantal varieties]{Schubert
varieties, toric varieties, and ladder determinantal varieties}
\author{N. Gonciulea}
\address{Department of Mathematics\\ Northeastern University\\ Boston, MA 02115}
\email{ngonciul@@lynx.neu.edu}
\author[V. Lakshmibai]{V. Lakshmibai${}^{\dag}$}

\email{ngonciul@@lynx.neu.edu, lakshmibai@@neu.edu}
\thanks{${}^{\dag}$Partially supported by NSF Grant DMS 9502942\\
\indent ${}^{\dag}$Partially suported by Northeastern University RSDF 95-96.}
\maketitle
\begin{abstract}
We contruct certain normal toric varieties (associated to finite distributive lattices) which are
degenerations of the Grassmannians. We also determine the singular loci for certain normal toric
varieties, namely the ones which are certain ladder determinantal varieties. As a consequence,
we  prove the conjecture of \cite{LS} on the components of the singular locus, for certain
Schubert varieties in the flag variety.
\end{abstract}


\section*{Introduction}
The results of this paper are two-folds. On the one  hand, we give a self-contained exposition
of the results of \cite{GLdef} on the degenerations of the Grassmannians to normal toric
varieties. On the other hand, we determine the irreducible components of the singular locus of the
determinantal variety (in an one-sided ladder) defined by the vanishing of all $ 2$
minors. Using the latter result, we prove the conjecture in \cite{LS} on the irreducible
components of the singular locus for certain Schubert varieties in the flag variety $SL(n)/B$.
Let $G_{d,n}$ be the Grassmannian of $d$-planes in $k^n$, $k$ being the base field, which is
assumed to be algebraically closed of arbitrary charcateristic. For $G_{d,n}$, as well as its
Schubert varieties, we proved in \cite{GLdef} their degenerations to normal toric varieties. The
normality of these toric varieties was concluded in \cite{GLdef} using the results of \cite{Hi}.
In this paper, among other things, we give an independent proof of normality of these toric
varieties, thus leading to a self-contained exposition of the results of \cite{GLdef} for
$G_{d,n}$. This is accomplished by showing that these varieties are Cohen-Macaulay, and are
nonsingular in codimension $1$ (using the Jacobian Criterion for smoothness, cf. Sections
\ref{s7} and \ref{s10}). We also prove some partial results (cf. Section \ref{s11}) towards the
determination of the irreducible components of the singular loci of these varieties, and state a
conjecture on the components of the singular loci. For the case $d=2$, these toric varieties are
in fact the determinantal varieties (in some  one-sided ladders) defined by the vanishing
of all the
$ 2$ minors. We determine (cf. Section \ref{s14}) the irreducible components of the
singular locus of a ladder determinantal variety $X(L)$ defined by the vanishing of all the
$ 2$ minors in  a general one-sided ladder $L$; as a consequence, we obtain an explicit
description of the irreducible components of the singular loci of the toric varieties considered
above for the case $d=2$ (and our conjecture is verified to be true in these cases). Further,
for a suitable $n$, and a suitable  parabolic subgroup $Q\subset SL(n)$, the variety $X(L)\times
\Bbb{A}^m$, where $m=\text{dim\,}SL(n)/Q-\# L$, gets identified with the ``opposite big cell" in
a certain Schubert variety $X(w)$in $SL(n)/Q$. The above mentioned result on the irreducible
components of the singular loci of these varieties enables us to verify the conjecture in
\cite{LS} for the associated Schubert varieties (see Section
\ref{s16} for details).
\begin{figure}
$$
\setcounter{MaxMatrixCols}{15}
\begin{matrix}

\ast&\ast&\dots&\ast&\a_1\\
\ast&\ast&\dots&\ast&\ast\\
\vdots&\vdots&&\vdots&\vdots\\
\ast&\ast&\dots&\ast&\ast\\
\ast&\ast&\dots&\ast&\ast&\ast&\dots&\ast&\a_2\\
\ast&\ast&\dots&\ast&\ast&\ast&\dots&\ast&\ast\\
\vdots&\vdots&&\vdots&\vdots&\vdots&&\vdots&\vdots\\
\ast&\ast&\dots&\ast&\ast&\ast&\dots&\ast&\ast\\
\vdots&\vdots&&\vdots&\vdots&\vdots&&\vdots&\vdots&\vdots\\
\vdots&\vdots&&\vdots&\vdots&\vdots&&\vdots&\vdots&\vdots&\vdots\\
\ast&\ast&\cdots&\ast&\ast&\ast&\cdots&\ast&\ast&\ast&\ast&\cdots&\cdots&\ast&\a_l\\
\ast&\ast&\cdots&\ast&\ast&\ast&\cdots&\ast&\ast&\ast&\ast&\cdots&\cdots&\ast&\ast\\
\vdots&\vdots&&\vdots&\vdots&\vdots&&\vdots&\vdots&\vdots&\vdots&&&\vdots&\vdots\\
\a_0&\ast&\cdots&\ast&\ast&\ast&\cdots&\ast&\ast&\ast&\ast&\cdots&\cdots&\ast&\ast
\end{matrix}
$$
\caption{An one-sided ladder}
\end{figure}


We now describe how the two-folds of the results of this paper are carried out. Let $\L$ be a
finite distributive lattice, $k[\L]=k[x_\t\mid\t\in\L]$, and $I(\L)$ the ideal generated by all
binomials of the form $x_\t x_\f -x_{\t\v\f}x_{\t\w\f}$, $\t$, $\f$ being two noncomparable
elements of $\L$ (see Section \ref{s3} for notations).  We first prove (cf. Section \ref{s3}) that
$I(\L)$  is a toric ideal (in the sense of
\cite{St}), and hence prime. We then give a short geometric proof (cf. Section \ref{s4}) of the
result that a binomial prime ideal is a toric ideal (in the sense of \cite{St}; an algebraic
proof is available in \cite{ES}). We next prove (cf. Section \ref{s7}) the Cohen-Macaulayness of
$R(\L)=k[\L]/I(\L)$. For the case $\L=I_{d,n}$, where $I_{d,n}=\{\ui=(i_1,\dots,i_d)\mid 1\le
i_1<\dots<i_d\le n\}$, $R(\L)$ is realized in \cite{GLdef}  as the special fiber of a certain
flat family whose general fiber is $k[G_{d,n}]$, the homogeneous coordinate ring of the
Grassmannian $G_{d,n}$ for the Pl\" ucker embedding. We prove (cf. Section \ref{s10}) that the
variety
$X(\L)=\text{Spec}R(\L)$ is nonsingular in codimension $1$. (This is accomplished by using the
Jacobian criterion for smoothness). Thus we obtain the normality of $X_{d,n}$. As a consequence,
we obtain (see Theorem \ref{10.8}):
\begin{thma}
The Grassmannian $G_{d,n}$ degenerates to the normal toric variety  $X_{d,n}$.
\end{thma}

Regarding the other half of the results of this paper, let $L$ be an one-sided ladder, and $X(L)$
the determinantal variety defined by the vanishing of all the $ 2$ minors in $L$. Let
$\a_1,\dots,\a_l$ be all the upper outer corners of $L$, and $\a_0$ the unique lower outer
corner of $\L$ (see Figure $1$). For
$1\le i\le l$, let
$V_i$ be the subvariety of $X(L)$ defined by $x_\a=0$ for $\a\in[\a_0\w\a_i,\a_0\v\a_i]$ (see
Section \ref{s2} for notations).

In Section \ref{s14} we prove the following result (see Theorem \ref{14.3}):
\begin{thmb}
The irreducible components of the singular locus of $X(L)$ are presisely $V_i$, $1\le i\le l$.
\end{thmb}
\noindent This is again proved using the Jacobian Criterion.

As mentioned above, $X(L)\times \Bbb{A}^m$ gets identified with the ``opoosite
 cell" in $X(w)$, for a certain Schubert variety $X(w)$ in $G/Q$, $G=SL(n)$, for a suitable $n$
and a certain parabolic subgroup $Q$.  Moreover, the varieties $V_i\times \Bbb{A}^m$, $1\le i\le
l$, also get identified with certain Schubert subvarieties $X(\te_i)$ of $X(w)$ (see Section
\ref{s15} for details). Let
$X(w^{\text{max}})=\pi^{-1}(X(w))$, where $\pi$ is the canonical projection $G/B\to G/Q$.
In Section \ref{s16} we prove the following result (see Theorem \ref{16.9}):
\begin{thmc}
The conjecture in \cite{LS} (on the irreducible components of the singular locus) holds for
$X(w^{\text{max}})$.
\end{thmc}
\noindent (see Section \ref{s16} for the statement of the conjecture).

The sections are organized as follows.  In Sections \ref{s1} and \ref{s2}, we recall some
generalities on Gr\" obner bases and distributive lattices. In Section \ref{s3}, we prove the
primality of the ideal $I(\L)$, $\L$ being any finite distributive lattice (our proof  is very
short, and combinatorial in nature). In Section \ref{s4}, we carry out a short geometric proof of
the fact that a binomial prime ideal is toric. In Section \ref{varlat}, we prove some general
results on
$X(\L)$ (in particular, we  compute the dimension of $X(\L)$).
In Section \ref{s6}, we construct a ``standard monomial basis" for $R(\L)$. In Section
\ref{s7}, we prove the Cohen-Macaulayness of $R(\L)$. In  Section
\ref{s8}, we derive some properties of the distributive lattice $I_{d,n}=\{\ui=(i_1,\dots,i_d)\mid
1\le i_1<\dots<i_d\le n\}$. In Section \ref{s9}, we study the variety $X_{d,n}$, where
$X_{d,n}=X(\L)$, for
$\L=I_{d,n}$. In Section \ref{s10}, we prove the normality of the variety $X_{d,n}$. In Section
\ref{s11}, we prove some partial results, and state a conjecture, on the irreducible components of
$\text{Sing\,}X_{d,n}$. In Section \ref{s12}, we recall the generalities on $G/Q$. In Section
\ref{s13}, we recall some results on the flag variety $SL(n)/B$, and its Schubert varieties.  In
Section \ref{s14}, we verify the conjecture stated in Section \ref{s11} for
$d=2$, by determining the irreducible components of $\text{Sing\,}X(L)$, where $X(L)$ is the
determinantal variety  defined  by the vanishing of all the
$ 2$ minors in a general one-sided ladder $L$. In Section \ref{s15}, we relate $X(L)$, as
well as the irreducible components of
$\text{Sing\,}X(L)$ to opposite  cells in certain Schubert varieties in certain $SL(n)/Q$.
In Section \ref{s16}, we verify the conjecture in \cite{LS} to be true for the pull-backs in $G/B$
(under $G/B\to G/Q$) of the Schubert varieties obtained in Section \ref{s15}.

It should be remarked that in \cite{Wa} a purely combinatorial description of
$\text{Sing\,}X(\L)$ is obtained. In this paper we have taken a geometric approach to this
problem, in the case when $\L=I_{d,n}$, or $\L$ is an one-sided ladder.

In this paper we have considered only those ladder determinantal varieties which are also toric
varieties. In a subsequent paper \cite{GLnext}, we study a larger class of ladder determinantal
varieties (which are not necessarily toric varieties), and prove results similar to those of
Sections \ref{s14}, \ref{s15} and \ref{s16}.

The authors wish to thank C. S. Seshadri and P. Sankaran for many useful discussions, especially
pertaining Sections \ref{s4} and \ref{s9}.



%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%% grobner bases

\section{Generalities on Gr\" obner bases}\label{s1}
Let $k$ be a field, and
cons{id}er the ring  $k[x_1,\ldots,x_n]$ of polynomials in $n$ variables
$x_1,\dots,x_n$. We recall below some generalities on Gr\" obner bases; for more
details one may refer to \cite{Cox}, \cite{Eisenbud}.
\begin{defn} A total order $\preceq$ on the set of monomials in
$k[x_1,\ldots,x_n]$ is called a\/ {\em monomial order} if for given monomials
$\mn$, $\mn_1$, $\mn_2$, with $\mn\ne 1$,
$\mn_1\prec\mn_2$ implies  $\mn_1\prec\mn\cdot \mn_1\prec\mn\cdot\mn_2$.
\end{defn}

For the rest of this Section, a fixed monomial order $\preceq$ is considered.

 If $f$
is a nonzero polynomial in $k[x_1,\dots,x_n]$, then the greatest monomial
(with respect to $\preceq$) occuring in $f$ is called the\/
{\em initial monomial of}  $f$, and we denote it by ${\rm in}(f)$;
 the coefficient of ${\rm in}(f)$ is called the initial coefficient of $f$.
 For a family of polynomials
 $\cal{F}\subset k[x_1,\dots,x_n]$,
the ideal generated by its elements will be denoted by $\langle \cal{F} \rangle$,
and the set of the initial monomials of all polynomials in
$\cal{F}$ will be denoted by ${\rm in}(\cal{F})$.
\begin{defn} Let $I\subset k[x_1,\dots,x_n]$ be an ideal.
 A finite set of
polynomials $\cal{F}\subset I$ is called a\/ {\em Gr\"obner basis for} $I$
\/ {\em with respect to} $\preceq$ if
$\langle{\rm in}(\cal{F})\rangle=\langle{\rm in}(I)\rangle$.
\end{defn}

\begin{defn}
A\/ {\em reduced Gr\"obner basis for} $I$\/ {\em with respect to} $\preceq$
is a Gr\"obner basis $\cal{F}$ for $I$ with respect to $\preceq$
such that the initial coefficients of the  elements in $\cal{F}$ are all\/ $1$,
and for all $f\in \cal{F}$,
none of the monomials present in $f$ lies in $\langle{\rm in}(\cal{F}
\setminus \{f\})\rangle$.
\end{defn}
\begin{prop}
 Any Gr\"obner basis for $I$  generates  $I$ as an ideal.
\end{prop}

In the case when $I$ is the defining ideal of an algebraic
variety $X$, a Gr\"obner basis for $I$ will be also called a {\it Gr\"obner
basis for $X$}.

\begin{prop}
A nonzero ideal
$I\subset k[x_1,\dots,x_n]$ has a unique reduced Gr\"obner
basis (with respect to a given monomial order).
\end{prop}
\subsection{Reverse lexicographic order}\label{rlex}
Assume that the variables
$x_1,\dots,x_n$ are toatally ordered as follows: $x_1<\dots <x_n$.
A monomial $\mn$ of degree $r$ in the polynomial ring $k[x_1,\dots,x_n]$
will be written in the form $\mn =x_{i_1}\dots x_{i_r}$, with
$1\le i_1\le \dots \le i_r\le n$. The {\em reverse lexicographic order\/} on the
set of monomials $\mn\in k[x_1,\dots,x_n]$ is denoted by $\preceq_{rlex}$,
and defined as follows: $x_{i_1}\dots x_{i_r}\prec_{rlex}
x_{j_1}\dots x_{j_s}$ if and only if either $r<s$, or $r=s$ and there
exists an $l<r$ such that $i_1=j_1,\dots,i_l=j_l,i_{l+1}<j_{l+1}$.
It is easy to check that $\preceq_{rlex}$ is a monomial oder.

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%% lattices
\section{Generalities on distributive lattices}\label{s2}
\begin{defn} A {\em lattice\/} is a partially ordered set $(\cal{L},\le)$ such that,
for every pair of elements $x,y\in \cal{L}$, there exist elements $x\v y$ and
$x\w y$, called the\/ {\em join}, respectively the\/ {\em meet} of $x$ and
$y$, defined by:
\begin{gather}
x\v y\ge x,\ x\v y\ge y,\text{ and if  }z\ge x \text{ and }
z\ge y,\text{ then } z\ge x\v y,\notag\\
x\w y\le x,\ x\w y\le y,\text{ and if  }z\le x \text{ and }
z\le y,\text{ then } z\le x\w y.\notag
\end{gather}
\end{defn}
It is easy to check that the operations $\v$ and $\w$ are commutative and
associative.
\begin{defn}
An element $z\in \cal{L}$ is called the\/ {\em zero} of $\cal{L}$, denoted by\/ $\widehat{0}$,
if $z\le x$ for all $x$ in $\cal{L}$. An element $z\in \cal{L}$ is called the\/
{\em one} of $\cal{L}$, denoted by\/ $\widehat{1}$, if $z\ge x$ for all $x$ in $\cal{L}$.
\end{defn}
\begin{defn}
Given a lattice $\cal{L}$, a subset $\cal{L}'\subset \cal{L}$ is called a\/
{\em sublattice} of $\cal{L}$ if $x,y\in\cal{L}'$ implies $x\w y\in\cal{L}'$,
$x\v y\in\cal{L}'$.
\end{defn}
\begin{defn}
Two lattices $\cal{L}_1$ and $\cal{L}_2$ are\/ {\em isomorphic} if there exists
a bijection $\varphi:\cal{L}_1\to \cal{L}_2$ such that, for all
$x,y\in\cal{L}_1$,
\begin{equation*}
\varphi(x\v y)=\varphi(x)\v \varphi(y) \text{   and   }
\varphi(x\w y)=\varphi(x)\w \varphi(y).
\end{equation*}
\end{defn}
\begin{defn} A lattice is called\/ {\em distributive} if the following
identities hold:
\begin{align}
x\w (y\v z)&=(x\w y)\v (x\w z)\\
x\v (y\w z)&=(x\v y)\w (x\v z).
\end{align}
\end{defn}
\subsection{An  example}
Given an integer $n\ge 1$, $\cal{C}(n)$ will denote the chain $\{1<\dots<n\}$,
and for $n_1,\dots,n_d>1$, $\cal{C}(n_1,\dots,n_d)$ will denote the chain
product lattice $\cal{C}(n_1)\times\dots\times\cal{C}(n_d)$
consisting of all $d$-tuples $(i_1,\dots,i_d)$, with $1\le i_1\le n_1,
\dots, 1\le i_d\le n_d$.
For $(i_1,\dots,i_d)$, $(j_1,\dots,j_d)$ in $\cal{C}(n_1,\dots,n_d)$,
we define
\begin{equation*}
(i_1,\dots,i_d)\le (j_1,\dots,j_d)\iff i_1\le j_1,\dots,
i_d\le j_d\ .
\end{equation*}
We have
\begin{gather}
\begin{split}
(i_1,\dots,i_d)\v (j_1,\dots,j_d)&=
(\max\{i_1,j_1\},\dots,\max\{i_d,j_d\})\notag\\
(i_1,\dots,i_d)\w (j_1,\dots,j_d)&=
(\min\{i_1,j_1\},\dots,\min\{i_d,j_d\}).\notag
\end{split}
\end{gather}

$\cal{C}(n_1,\dots,n_d)$ is a finite distributive lattice, and its zero and one
are $(1,\dots,1)$, $(n_1,\dots,n_d)$  respectively.

Note that there is a total order $\lt$ on $\cal{C}(n_1,\dots,n_d)$ extending $<$,
namely the lexicographic order, defined by
$(i_1,\dots,i_d)\lt (j_1,\dots,j_d)$ if and only if there exists $l<d$
such that $i_1=j_1,\dots,i_l=j_l,i_{l+1}<j_{l+1}$. Also note that two elements
$(i_1,\dots,i_d)\lt (j_1,\dots,j_d)$ are non-comparable with respect to
$\le$ if and only if there exists $1<h\le d$ such that $i_h>j_h$.

Sometimes we denote the elements of $\cal{C}(n_1,n_2,\dots,n_d)$ by
$x_{i_1\dots i_d}$, with
$1\le i_1\le n_1, \dots, 1\le i_d \le n_d$.

\subsection{} The lattice of all subsets of the set $\{1,2,\dots,n\}$ is denoted
by $\cal{B}(n)$, and called the {\em Boolean algebra of rank\/} $n$. Note that
$\cal{B}(n)$ is isomorphic to $[\cal{C}(2)]^n$.

One has the following (see \cite{Aigner}):
\begin{thm} \label{distlat}
Any finite distributive lattice  is isomorphic to a sublattice of a
Boolean algebra of finite rank, and hence, in particular, to a sublattice of a
finite chain product.
\end{thm}



%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%% The ideal associated to a  distrubutive lattice is toric

\section{The ideal  associated to a distributive lattice}\label{s3}
\subsection{}Let $T=(k^*)^m $ be the $m$-dimensional torus. Let $M$ be the
character group (=Hom$_{{\rm alg. gp.}}(T,\Bbb{G}_m)$) of $T$. Then $M$ can be
identified with $\Bbb{Z}^m$.
Let $\cal{A}=\{\bold{a}_1,\dots,\bold{a}_n\}$ be a subset of $\Bbb{Z}^m$. Consider the
homomorphism
\begin{equation*}
\pi:\Bbb{Z}^n\to \Bbb{Z}^m,\quad \bold{u}=(u_1,\dots,u_n)\mapsto
u_1\bold{a}_1+\dots+u_N\bold{a}_N.
\end{equation*}
Let $\bold{x}=(x_1,\dots,x_n)$, $\bold{t}=(t_1,\dots,t_m)$, and
\begin{equation*}
k[\bold{x}]=k[x_1,\dots,x_n],\quad k[\bold{t}^{\pm 1}]=k[t_1,\dots,t_m,
t_1^{-1},\dots,t_m^{-1}].
\end{equation*}
 The map $\pi$ induces  a homomorphism of semigroup algebras
\begin{equation*}
\widehat{\pi}:k[\bold{x}]\to k[\bold{t}^{\pm 1}],\qquad x_i\mapsto\bold{t}^
{\bold{a}_i}.
\end{equation*}

\begin{defn} \label{3.2}
The kernel of $\widehat{\pi}$ is denoted by $I_{\cal{A}}$
and called the\/ {\em toric ideal} associated to $\cal{A}$.
\end{defn}

Note that a toric ideal is prime.

Recall the following (see \cite{St}).
\begin{prop} \label{vegen}
The toric ideal $I_{\cal{A}}$ is spanned as a a $k$-vector space by the set of
binomials
\begin{equation}\label{veset}
\{\bold{x}^{\bold{u}}-\bold{x}^{\bold{v}}\mid \bold{u},\bold{v}\in \Bbb{Z_+}^n
\text{ with } \pi (\bold{u})=\pi(\bold{v})\}.
\end{equation}
\end{prop}

\subsection{An important example}
Let us fix the integers $n_1,\dots,n_d>1$, and let
$n=\Pi_{i=1}^{d}n_i$, $m=\sum_{i=1}^{d}n_i$. Let
$\bold{e}^l_1,\dots,\bold{e}^l_{n_l}$ be the unit vectors in
$\Bbb{Z}^{n_l}$, for $1\le l\le d$. For $1\le \xi _1\le n_1,
\dots,1\le \xi _d\le n_d$, define
\begin{equation*}
\bold{a}_{\xi _1\dots \xi _d}=\bold{e}^1_{\xi _1}\oplus
\dots\oplus\bold{e}^d_{\xi _d}\in \Bbb{Z}^{n_1}\oplus\dots\oplus \Bbb{Z}^{n_d}
\end{equation*}
and
\begin{equation*}
\cal{A}_{n_1,\dots,n_d}=\{\bold{a}_{\xi _1\dots \xi _d}\mid 1\le \xi _1\le n_1,
\dots,1\le \xi _d\le n_d\}.
\end{equation*}
The corresponding map
\begin{equation*}
\pi:\Bbb{Z}^{n_1\cdot\dots\cdot n_d}\to\Bbb{Z}^{n_1+\dots +n_d}
\end{equation*}
is defined as follows: for $1\le l\le d$ and $1\le i_l\le n_l$ fixed, the
$(n_1+\dots +n_{l-1}+i_l)$-th coordinate of $\pi(\bold{u})$ is given by
$\sum u_{\xi _1\dots \xi _{l-1}\xi _l\xi _{l+1}\dots \xi _d}$, the sum being
taken over the elements
$(\xi_1,\dots,\xi_{l-1},\xi_l,\xi_{l+1},\dots,\xi_d)$ of
$\cal{C}(n_1,\dots,n_d)$ with $\xi_l=i_l$. We call this subset the
$l$-th {\em  slice} of $\cal{C}(n_1,\dots,n_d)$ defined by $i_l$, and denote it
by $\{\xi _l=i_l\}$.
The components (or {\em  entries\/}) of an element
$\bold{u} \in \Bbb{Z}^{n_1\cdot\dots\cdot n_d}$  are indexed by the
elements $(i_1,\dots,i_d)$ of $\cal{C}(n_1,\dots,n_d)$.
If $(j_1,\dots, j_d)\in \{\xi_l=i_l\}$, sometimes we also say that
$u_{j_1\dots j_d}$ itself belongs to the slice $\{\xi _l=i_l\}$.

The map $\pi$ induces the map
\begin{equation*}
\widehat{\pi}:k[x_{11\dots 1},\dots,x_{\xi _1\xi _2\dots \xi _d},\dots,
x_{n_1n_2\dots n_d}]\to k[t_{11},\dots,t_{1n_1},\dots,t_{d1},\dots,t_{dn_d}]
\end{equation*}
given by
\begin{equation*}
x_{\xi _1\dots \xi _d}\mapsto t_{1\xi _1}\dots t_{d\xi _d},\text{  for  }
1\le \xi _1\le n_1, \dots,1\le \xi _d\le n_d.
\end{equation*}


\begin{defn}\label{Id}
Given a finite lattice $\cal{L}$, the\/ {\em ideal  associated to}
$\cal{L}$, denoted by $I(\cal{L})$, is
the ideal of the polynomial ring $k[\cal{L}]$ generated by the set of
binomials
\begin{equation*}
\cal{G}_{\cal{L}}=\{xy-(x\w y)(x\v y)\mid x,y\in\cal{L} \text{ non-comparable}\}.
\end{equation*}
\end{defn}

By Theorem \ref{distlat}, a finite distributive lattice $\cal{L}$  may be
identified with a sublattice of a finite chain product lattice . Hence it inherits a total
order extending the given partial order.
In turn, this total order induces the reverse lexicographic order on the monomials
in $k[\cal{L}]$, as in \ref{rlex}.

The following theorem shows that the ideal associated to a chain product lattice
is toric.
\begin{thm} \label{chaintor}
$1)$ We have $I(\cal{C}(n_1,\dots,n_d))=I_{\cal{A}_{n_1,\dots,n_d}}$.

$2)$ The set of binomials
\begin{equation*}
\cal{G}=\{xy-(x\w y)(x\v y)\mid x,y\in\cal{C}(n_1,\dots,n_d) \text{ non-comparable}\}
\end{equation*}
is a Gr\" obner basis for $I(\cal{C}(n_1,\dots,n_d))$ with respect to the
reverse lexicographic order.
\end{thm}
\begin{pf}
Let $\C=\cal{C}(n_1,\dots,n_d)$ and $\A=\cal{A}_{n_1,\dots,n_d}$.
Let $f\in I_{\cal{A}}$; by Proposition \ref{vegen}, there exist $\bold{u}_i,
\bold{v}_i\in\Bbb{Z_+}^n$ with $\pi(\bold{u}_i)=\pi(\bold{v}_i)$, and $c_i\in k^*$,
$1\le i\le s$ such that
\begin{equation}\label{zzero}
f=\sum_{i=1}^s c_i(\bold{x}^{\bold{u}_i}-\bold{x}^{\bold{v}_i})
\end{equation}
for some $s\ge 1$, with the property that $s$ is the smallest integer $\ge 1$
such that $f$ can be expressed as a linear combination of $s$ binomials in
the set (\ref{veset}). Now we rewrite $f$ as
\begin{equation*}
f=\sum_{i=1}^s a_i \bold{x}^{\bold{u}_i}+\sum_{i=1}^s b_i \bold{x}^{\bold{v}_i},
\quad a_i,b_i\in k.
\end{equation*}
Then none of the coefficients $a_1,\dots,a_s,b_1,\dots,b_s$ is zero. Indeed,
suppose that $a_i=0$ for some $1\le i\le s$. This implies that
 there exists $j\in\{1,\dots,s\}$, $j\ne i$ such that
either $c_j=c_i$ and $\bold{v}_j=\bold{u}_i$, or $c_j=-c_i$ and
$\bold{u}_j=\bold{u}_i$. In the first case we have
\begin{equation} \label{first}
c_i(\bold{x}^{\bold{u}_i}-\bold{x}^{\bold{v}_i})+
c_j(\bold{x}^{\bold{u}_j}-\bold{x}^{\bold{v}_j})=
c_i(\bold{x}^{\bold{u}_j}-\bold{x}^{\bold{v}_i}),\quad
\pi(\bold{u}_j)=\pi(\bold{v}_i).
\end{equation}
In the second case we have
\begin{equation} \label{second}
c_i(\bold{x}^{\bold{u}_i}-\bold{x}^{\bold{v}_i})+
c_j(\bold{x}^{\bold{u}_j}-\bold{x}^{\bold{v}_j})=
c_i(\bold{x}^{\bold{v}_j}-\bold{x}^{\bold{v}_i}),\quad
\pi(\bold{v}_j)=\pi(\bold{v}_i).
\end{equation}
But (\ref{zzero}), (\ref{first}) and (\ref{second}) imply that $f$ can be written
as a linear combination of $s-1$ binomials in the set (\ref{veset}),
contradicting the choice of $s$. Thus $a_i\ne 0$, $1\le i\le s$, and similarly
$b_i\ne 0$, $1\le i\le s$. This shows that ${\rm in}(f)={\rm in}
(\bold{x}^{\bold{u}_i}-\bold{x}^{\bold{v}_i})$ for some $1\le i\le s$. Let us
write
\begin{equation*}
\bold{x}^{\bold{u}_i}-\bold{x}^{\bold{v}_i}=\bold{x}^{\bold{w}}
(\bold{x}^{\bold{u}}-\bold{x}^{\bold{v}}),
\end{equation*}
where $\bold{u},\bold{v},\bold{w}\in \Bbb{Z_+}^n$, with $\pi (\bold{u})=\pi(\bold{v})$ and
${\rm supp}(\bold{u})\cap {\rm supp}(\bold{v})=\emptyset$.
Let us suppose that $\bold{x}^{\bold{u}}\succeq_{rlex}\bold{x}^{\bold{v}}$,
i.e. ${\rm in}(\bold{x}^{\bold{u}}-\bold{x}^{\bold{v}})=\bold{x}^{\bold{u}}$ and
${\rm in}(f)={\rm in}(\bold{x}^{\bold{u}_i}-\bold{x}^{\bold{v}_i})=
\bold{x}^{\bold{u}_i}$.
Let
$x_{i_1\dots i_d}$ be the smallest variable appearing in $\bold{x}^{\bold{u}}$,
i.e. $(i_1,\dots,i_d)$ is the smallest element of
${\rm supp}(\bold{u})$ with respect to $\lt$.
Then $\bold{x}^{\bold{v}}$ contains a variable $x_{k_1\dots k_d}$, with
$(k_1,\dots,k_d)\lt (i_1,\dots,i_d)$.
Since $\pi (\bold{u})=\pi(\bold{v})$, the sum of the entries
in every slice is the same for both $\bold{u}$ and $\bold{v}$. In particular,
since all the entries of $\bold{u}$ in the slices $\{\xi_1=i\}$, with
$1\le i<i_1$, are $0$ (by the choice of $(i_1,\dots,i_d)$), all the entries of
$\bold{v}$ in these slices must also be $0$. This implies that
$(k_1,\dots,k_d)\in \{\xi_1=i_1\}$.
Let $1<h\le n_1$ such that $k_1=i_1,\dots,k_{h-1}=i_{h-1},k_h<i_h$.
Then the  sum of the elements of $\bold{v}$ in the slice $\{\xi_h=k_h\}$ is
nonzero, which implies that $\{\xi_h=k_h\}\cap {\rm supp}(\bold{u})\ne\emptyset$.
Let $(j_1,\dots,j_d)$ be an element in this intersection. We have
$(i_1,\dots i_d)\lt (j_1,\dots,j_d)$ (by the definition of
$(i_1,\dots i_d)$), and since $i_h>k_h=j_h$, we conclude that
$(i_1,i_2,\dots,i_d)$ and $(j_1,\dots,j_d)$ are non-comparable.
Thus we obtain  that
$\bold{x}^{\bold{u}}$ is divisible by $x_{i_1\dots i_d}x_{j_1\dots j_d}$.
Hence $\bold{x}^{\bold{u}_i}$ is also divisible by
$x_{i_1\dots i_d}x_{j_1\dots j_d}$.
Therefore ${\rm in}(f)$ is divisible by the initial term of an element
of the set
\begin{equation*}
\cal{G}=\{xy-(x\w y)(x\v y)\mid x,y\in\C \text{ non-comparable}\}
\end{equation*}
of generators of the ideal $I(\C)$. Since $\cal{G}\subset I_{\A}$, it follows
that $\cal{G}$ is a Gr\" obner basis for $I_{\A}$. In particular it is a set
of generators for this ideal. Thus $\cal{G}$ generates both $I(\C)$ and
$I_{\A}$, which  implies the equality of the two ideals.
\end{pf}


\begin{thm} \label{3.7}
Let $\cal{L}$ be a finite distributive lattice . Then

$1)$ The ideal $I(\cal{L})$  is toric (and hence prime).

$2)$ The set of binomials
\begin{equation*}
\cal{G}_{\cal{L}}=\{xy-(x\w y)(x\v y)\mid x,y\in\cal{L} \text{ non-comparable}\}
\end{equation*}
is a Gr\" obner basis for $I(\cal{L})$ with respect to the reverse lexicographic
order.
\end{thm}
\begin{pf}
By Theorem \ref{distlat}, we may assume that $\cal{L}$ is a
sublattice of $\cal{C}(n_1,\dots,n_d)$, for some $n_1,\dots,n_d\ge 1$.
Let us denote  $\C=\cal{C}(n_1,\dots,n_d)$, $\A=\cal{A}_{n_1,\dots,n_d}$ and
$\cal{G}=\cal{G}_{\C}$. Note that $\cal{G}_{\cal{L}}$ is the subset of $\cal{G}$
consisting of all binomials in $\cal{G}$ involving only the variables from $\cal{L}$.
Let us denote
\begin{equation*}
\cal{G}_{\cal{L}}=\{f_1,\dots,f_r\},\quad \cal{G}\setminus \cal{G}_{\cal{L}}=
\{g_1,\dots,g_s\}
\end{equation*}
Let $g_i=xy-(x\w y)(x\v y)$, with $x,y\in\C$ non-comparable, $1\le i\le s$;
then at least one of $x$ and $y$ does not belong to $\cal{L}$ ($\cal{L}$ being
a sublattice of $\C$, $x,y\in \cal{L}$ would imply $x\w y, x\v y\in \cal{L}$,
so $g_i$ would involve only variables from $\cal{L}$).

Let $\A_{\cal{L}}\subset\A$ be given by the elements in $\A$ indexed by the
elements of
$\cal{L}$, and let $f$ be an element of
\begin{equation*}
I_{\A_{\cal{L}}}=\ker \left( \widehat{\pi}\bigr|_{k[\cal{L}]}\right) =
(\ker \widehat{\pi})\cap k[\cal{L}]=I_{\A}\cap k[\cal{L}].
\end{equation*}
In the course of the proof of Theorem \ref{chaintor}, we saw that ${\rm in}(f)$
is divisible by the initial term of a binomial in $\cal{G}$, and since
$f\in k[\cal{L}]$, this binomial must be one of the $f_i$'s, i.e. an element of
$\cal{G}_{\cal{L}}$. Since
$\cal{G}_{\cal{L}}\subset I_{\A_{\cal{L}}}$, it follows that $\cal{G}_{\cal{L}}$
is a Gr\" obner basis for $I_{\A_{\cal{L}}}$, hence $\cal{G}_{\cal{L}}$
generates $I_{\A_{\cal{L}}}$. Therefore
$I(\cal{L})=I_{\A_{\cal{L}}}$, and the stated assertions follow now.
\end{pf}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%% binomial varieties

\section{Varieties defined by binomials}\label{varbino}\label{s4}
Let $k$ be  an algebraically closed field of arbitrary characteristic. Consider an integer
$m\ge 1$, and the $m$-dimensional torus $T_m=(k^*)^m$
\begin{defn} (cf. \cite{KK})
{\em An equivariant affine embedding of} $T_m$ is an affine variety $X$ containing $T_m$ as an
open subset, equipped with an action of $T_m$ on $X$ $$T_m\times X\to X$$ extending the
action $T_m\times T_m\to T_m$ given by the translations in $T_m$.
\end{defn}
\begin{defn} (cf. \cite{F}, \cite{KK})
An equivariant affine embedding $X$ of a torus is called an \/ {\em affine toric variety}
if it is normal.
\end{defn}

\subsection{}
Let $N\ge 1$, and let $X$ be an affine variety in $\Bbb{A}^N$, not contained in any of the
coordinate hyperplanes $\{ x_i=0\}$. Further, let $X$ be irreducible, and let its defining prime
ideal $I(X)$ be generated by $l$ binomials
\begin{equation*}
x_1^{a_{i1}}\dots x_N^{a_{iN}}-\l_i x_1^{b_{i1}}\dots x_N^{b_{iN}}\quad ,\qquad 1\le i\le l\quad
.\tag{$*$}
\end{equation*}
Consider the natural action of the torus $T_N=(k^*)^N$ on $\Bbb{A}^N$,
\begin{equation*}
(t_1,\dots,t_N)\cdot (a_1,\dots,a_N)=(t_1a_1,\dots,t_Na_N).
\end{equation*}
Let $X(T_N)=\text{Hom} (T_N,\Bbb{G}_m)$ be the character group of $T_N$, and let
$\e_i\in X(T_N)$ be the character
\begin{equation*}
\e_i(t_1,\dots,t_N)=t_i\quad ,\qquad 1\le i\le N.
\end{equation*}
For $1\le i\le l$, let
\begin{equation*}
\varphi_i=\sum_{t=1}^N(a_{it}- b_{it})\e_t\quad .
\end{equation*}
Set $T=\cap _{i=1}^l\ker \varphi_i$, and $X^{\circ}=\{(x_1,\dots,x_N)\in X \mid x_i\ne 0 \text{ for all
} 1\le i\le N\}$.

\begin{prop} \label{binotoric}
Let notations be as above.

$(1)$ There is a canonical action of $T$ on $X$.

$(2)$  $X^{\circ}$ is $T$-stable. Further, the action of $T$ on $X^{\circ}$ is simple and
transitive.

$(3)$ $T$ is a subtorus of $T_N$, and $X$ is an equivariant affine embedding of $T$.
\end{prop}
\begin{pf}
$(1)$ We consider  the  (obvious) action of $T$ on $\Bbb{A}^N$.
Let $(x_1,\dots,x_N)\in X$,  $\bold{t}=(t_1,\dots,t_N)\in T$, and $(y_1,\dots,y_N)=
\bold{t}\cdot (x_1,\dots,x_N)=(t_1x_1,\dots,t_Nx_N)$.
Using the fact that $(x_1,\dots,x_N)$ satisfies $(*)$, we
obtain
\begin{equation*}
y_1^{a_{i1}}\dots y_N^{a_{iN}}=t_1^{a_{i1}}\dots t_N^{a_{iN}}x_1^{a_{i1}}\dots x_N^{a_{iN}}=
\l_i t_1^{b_{i1}}\dots t_N^{b_{iN}}x_1^{b_{i1}}\dots x_N^{b_{iN}}=\l_i y_1^{b_{i1}}\dots
y_N^{b_{iN}},
\end{equation*}
for all $1\le i\le l$, i.e. $(y_1,\dots,y_N)\in X$.
Hence $\bold{t}\cdot (a_1,\dots,a_N)\in X$ for all $\bold{t}\in X$.

$(2)$ Let $x=(x_1,\dots,x_N)\in X^{\circ}$, and $\bold{t}=(t_1,\dots,t_N)\in T$. Then, clearly
$\bold{t}\cdot (x_1,\dots,x_N)\in X^{\circ}$. Considering $x$ as a point in
$\Bbb{A}^N$, the isotropy subgroup in $T_N$ at $x$ is  $\{\text{id}\}$. Hence the isotropy
subgroup in $T$ at $x$ is also $\{\text{id}\}$. Thus the action of $T$ on $X^{\circ}$ is simple.

Let $(x_1,\dots,x_N)$, $(x_1',\dots,x_N')\in X^{\circ}$. Set
$\bold{t}=(t_1,\dots,t_N)$, where $t_i=x_i/x_i'$. Then, clearly $\bold{t}\in T$. Thus
$(x_1,\dots,x_N)=\bold{t}\cdot (x_1',\dots,x_N')$. Hence the action of $T$ on $X^{\circ}$ is
simple and transitive.

$(3)$ Now, fixing a point $x\in X^{\circ}$, we obtain from $(2)$ that the orbit map
$\bold{t}\mapsto \bold{t}\cdot x$ is in fact an isomorphism of $T$ onto $X^{\circ}$. Also,
since $X$ is not contained in any of the coordinate hyperplanes, the open set
$X_i=\{(x_1,\dots,x_N)\in X\mid x_i\ne 0\}$ is nonempty for all $1\le i\le N$. The irreducibility
of $X$ implies that the sets $X_i$, $1\le i\le N$, are open dense in $X$, and hence their
intersection \begin{equation*} X^{\circ}=\bigcap_{i=1}^NX_i=\{(x_1,\dots,x_N)\in X \mid x_i\ne 0
\text{ for any $i$}\} \end{equation*} is an open dense set in $X$, and thus $X^{\circ}$ is
irreducible. This implies that $T$ is irreducible (and hence connected). Thus $T$ is a subtorus
of $T_N$. The assertion that $X$ is an equivariant affine embedding of $T$ follows from $(1)$
and $(2)$.
\end{pf}
\begin{rem}
One can see that the ideal $I(X)$ is a toric ideal in the sense of Definition \ref{3.2}; more
precisely,
$I(X)=I_{\cal{A}}$, with $m=\dim T$, $n=N$, $\cal{A}=\{\rho_i,1\le i\le N\}$,  where
$\rho_i=\e_i\bigr|_T$ (here
$k[T]$ is identified with $k[t_1^{\pm 1},\dots,t_m^{\pm 1}]$, and the character group
$X(T)$ with $\Bbb{Z}^m$).
\end{rem}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%% X(\L)

\section{Some general results on the variety associated to a finite distributive
lattice}\label{varlat}


We first recall some basic definitions on finite partially ordered sets.
A partially ordered set is also called a poset.
\begin{defn}
A finite poset $P$ is called\/ {\em bounded} if it has a unique maximal, and a
unique minimal element, denoted $\widehat{1}$ and $\widehat{0}$ respectively.
\end{defn}
\begin{defn}
A totally ordered subset $C$ of a finite poset $P$ is called a\/ {\em chain}, and the
number $\# C-1$ is called the\/ {\em length} of the chain.
\end{defn}
\begin{defn}
A bounded poset $P$ is said to be\/ {\em graded} if all maximal chains have the same
length (note that $\widehat{1}$ and $\widehat{0}$ belong to any maximal chain).
\end{defn}
\begin{defn}
Let $P$ be a graded poset. The length of a maximal chain in $P$ is called the\/
{\em rank} of $P$.
\end{defn}
\begin{defn}\label{5.5}
Let $P$ be a graded poset. For $\l,\m\in P$ with $\l\ge\m$, the graded poset
$\{\t\in P\mid\m\le\t\le\l\}$ is called the\/ {\em interval from $\m$ to $\l$}, and
denoted by $[\m,\l]$. The rank of $[\m,\l]$ is denoted by $l_{\m}(\l)$;
if $\m=\widehat{0}$, then  we denote $l_{\m}(\l)$ by just $l(\l)$.
\end{defn}
\begin{defn}
Let $P$ be a graded poset, and $\l,\m\in P$, with $\l\ge\m$. The ordered  pair $(\l,\m)$ is called
a\/ {\em cover} (and we also say that $\l$\/ {\em  covers} $\m$) if $l_{\m}(\l)=1$.
\end{defn}
\begin{defn}
An element $z$ of a lattice $\L$ is called\/ {\em join-irreducible} (respectively\/
{\em meet-irreducible}) if $z=x\v y$ (respectively $z=x\w y$) implies $z=x$ or $z=y$. The
set of join-irreducible (respectively meet-irreducible) elements of $\L$ is denoted
by $J_{\L}$ (respectively $M_{\L}$), or just by $J$ (respectively $M$) if
 no confusion is possible.
\end{defn}
\begin{defn}
The set $J_{\L}\cap M_{\L}$ of join and meet-irreducible elements is denoted by $JM_{\L}$, or just
$JM$ if no confusion is possible.
\end{defn}
\begin{defn}
A subset $I$ of a poset $P$ is called an {\em ideal} of $P$ if for all $x,\,y\in P$,
$$x\in I\text{ and }y\le x\text{ imply }y\in I.$$
\end{defn}
\begin{thm}\label{5.10} (Birkhoff)
Let $\L$ be a distributive lattice with $0$, and $P$ the poset of its nonzero join-irreducible
elements. Then $\L$ is isomorphic to the lattice of finite ideals of $P$, by means of the
lattice isomorphism
$$\a\mapsto I_{\a}=\{\t\in P\mid \t\le\a\},\qquad \a\in\L.$$
\end{thm}
\begin{defn}
A quadruple of the form $(\t,\f,\t\v\f,\t\w\f)$, with $\t,\f\in\L$ non-comparable
is called a\/ {\em diamond}, and is denoted by $D(\t,\f,\t\v\f,\t\w\f)$.
\end{defn}
\begin{lem}
With the notations as above, we have

$(a)$ $J=\{\t\in\L\mid\text{there exists at most one cover of the form } (\t,\l)\}$.

$(b)$ $M=\{\t\in\L\mid\text{there exists at most one cover of the form } (\l,\t)\}$.
\end{lem}
\begin{pf}
In order to prove part $(a)$, let us denote
\begin{equation*}
Z=\{\l\in\L\mid\text{there exists at most one cover of the form } (\l,\rho)\}
\end{equation*}
Clearly, $J\subset Z$. Let $\l\in Z$, and assume $\l\not\in J$. Assumption implies that  there
exists a diamond
$D(\t,\f,\l,\m)$. In particular, this implies $\t\le\l$, $\f\le\l$. Now  $\L$ being
graded, there exist $\a$, $\b$ such that $\t\le\a$, $\f\le\b$, and $(\l,\a)$, $(\l,\b)$
are covers. Now, $\t\le\a$, $\f\le\b$, and $\l=\a\v\b$ imply $\a\ne\b$. But this would
imply that $\l$ covers two distinct elements, contradicting the hypothesis that $\l\in Z$.
Hence the assumption that $\l\not\in J$ is wrong, which shows $(a)$.

The proof of $(b)$ is similar.
\end{pf}

\subsection{}\label{5.13}
Let $X(\L)$ be the affine variety of the zeroes in $k^N$ of $I(\L)$ (note that $X(\L)$ is
irreducible, in view of Theorem \ref{3.7}).
 Then $X(\L)$ is a variety defined by binomials, and we follow the
notations in Section \ref{varbino}. We have $N=\#\L$, $I(X(\L))=I(\L)$. Let
$\cal{I}=\{(\t,\f,\t\v\f,\t\w\f)\mid (\t,\f)\in Q\}$, where
\begin{equation*}
Q=\{(\t,\f)\mid \t,\f\in\L\text{ non-comparable}\}.
\end{equation*}
In view of Proposition \ref{binotoric}, $\dim X(\L)=\dim T$, and we now compute the
dimension of $T$.

Let $\pi:X(T_N)\to X(T)$ be the canonical map, given by restriction, and for $\h\in X(T_N)$,
denote $\pi (\h)$ by $\overline{\h}$. Let us fix a $\Bbb{Z}$-basis $\{\h_\t\mid\t\in\L\}$ for
$X(T_N)$. For a diamond $D=(\t,\f,\t\v\f,\t\w\f)\in\cal{I}$, let
$\h_D=\h_{\t\v\f}+\h_{\t\w\f}-\h_{\t}-\h_{\f}$.

\begin{lem}\label{prev}
We have

$(1)$ $X(T)\simeq X(T_N)/\ker\pi$.

$(2)$ $\ker\pi$ is generated by $\{\h_D\mid D\in\cal{I}\}$.
\end{lem}
\begin{pf}
The canonical map  $\pi$ is, in fact,  surjective, since $T$ is a subtorus of $T_N$. Now $(1)$
follows from this. The assertion $(2)$ follows from the definition of $T$.
\end{pf}

\subsection{}\label{5.15}
 For $\a\in\L$, let $I_\a$ be the
ideal corresponding to $\a$ under the isomporhism in Theorem \ref{5.10}.   Let
$$\psi_\a=\sum_{\te\in I_\a}\h_\te.$$

\begin{lem}\label{before}
The  set $\{\overline{\psi}_a\mid a\in\L\}$ generates $X(T)$ as a $\Bbb{Z}$-module.
\end{lem}
\begin{pf}
Consider the homomorphism
$$\te :X(T_N)\to X(T)\quad ,\quad \h_a\mapsto \overline{\psi}_a.$$
(note that $\{\h_a\mid a\in\L\}$ is a $\Bbb{Z}$-basis for $X(T_N)$). For a diamond
$D=(\t,\f,\m,\l)\in\cal{I}$, we have
\begin{equation*}
\overline{\psi}_\t+\overline{\psi}_\f=\overline{\psi}_\m+\overline{\psi}_\l,\tag{$*$}
\end{equation*}
and hence $\h_D\in\ker\te$. Conversely, it is clear that any relation among $\overline{\psi}_a$'s
is of the form $(*)$. Hence $\ker\te=\ker\pi$, and $X(T)\simeq X(T_N)/\ker\te$,
(cf. Lemma \ref{prev}). In particular, this implies that $\te$ is onto, and the result follows,
since $\{\h_a\mid a\in\L\}$ generates $X(T_N)$ as a $\Bbb{Z}$-module.
\end{pf}
\begin{prop}
The set $\{\overline{\h}_\t\mid \t\in J\}$ is a $\Bbb{Z}$-basis for $X(T)$.
\end{prop}
\begin{pf}
By Lemma \ref{before}, $\{\overline{\psi}_a\mid a\in\L\}$ generates the $\Bbb{Z}$-module
$X(T)$. Now
\begin{equation*}
\overline{\psi}_a=\sum_{\te\in I_a}\overline{\h}_\te=\sum_{\{\te\in
J_{\L},\te\le a\}}\overline{\h}_{\te}\tag{\dag}
\end{equation*}
\noindent (cf. \ref{5.15}). Hence
$\{\overline{\h}_\t\mid\t\in J\}$ generates the $\Bbb{Z}$-module $X(T)$. Also it is clear that
no proper subset of $\{\overline{\h}_\t\mid\t\in J\}$   generates $X(T)$. The result now follows.
\end{pf}
Now, since $\dim X(\L)=\dim T$, we obtain
\begin{thm}
The dimension of $X(\L)$ is equal to  $\#J_{\L}$.
\end{thm}
\begin{defn}
Let $\L$ be a finite distributive lattice. We call the cardinality of $J_{\L}$  the {\em
dimension} of $\L$, and we denote it by $\dim\L$. If $\L'$ is a sublattice of $\L$, then the
{\em codimension} of $\L'$ in $\L$ is defined as $\dim\L-\dim\L'$.
\end{defn}

\begin{lem}\label{5.20}
Let $P=(P_\te)_{\te\in\L}\in X(\L)$ be such that $P_\t\ne 0$ for any $\t\in J$. Then
$P_\te\ne 0$ for any $\te\in\L$.
\end{lem}
\begin{pf} (by induction) Let $\te\in\L$. If $\te\in J_{\L}$, there is nothing to check. Let
then $\te\in\L\setminus J_{\L}$.

Let $\te$ be a minimal element of $\L\setminus J_{\L}$. This
implies that every $\t\in \L$  such that $\t <\te$ belongs to $J_{\L}$. The fact that
$\te\in\L\setminus J_{\L}$ implies that there are at least two elements $\te_1,\te_2$ of $\L$
which are covered by $\te$. Note that $\te_1$, $\te_2$ are not comparable. We have
$\te_1\v\te_2=\te$. Let $\m=\te_1\w\te_2$. We have $P_\te P_\m =P_{\te_1}P_{\te_2}$. Now
$P_{\te_1}\ne 0$, $ P_{\te_2}\ne 0$, since $\te_1, \te_2\in J_{\L}$. Hence we obtain that
$P_{\te}\ne 0$.

Let now $\f$ be any element of $\L\setminus J_{\L}$. Assume, by induction,
that $P_\t\ne 0$ for any $\t<\f$.
Since
$\f\not\in J_{\L}$, there are at least two elements $\f_1,\f_2$ of $\L$ which are covered by $\f$. We have
$\f_1\v\f_2=\f$, $P_\f P_\d =P_{\f_1}P_{\f_2}$, where  $\d=\f_1\w\f_2$. Also $P_{\f_1}\ne 0$,
$P_{\f_2}\ne 0$ (since $\f_1$, $\f_2$ are both $<\f$). Hence we obtain $P_\f\ne 0$.
\end{pf}


\begin{defn}\label{5.21}(cf. \cite{Wa})
A sublattice  $\L '$ of $\L$ is called an {\em embedded sublattice of} $\L$ if
$$\t,\,\f\in\L,\quad\t\v\f,\,\t\w\f\in\L '\quad\Rightarrow\quad\t,\,\f\in\L '.$$
\end{defn}
Given a sublattice $\L '$ of $\L$, let us consider the variety $X(\L ')$, and consider the
canonical embedding $X(\L ')\hookrightarrow\Bbb{A}(\L ')\hookrightarrow\Bbb{A}(\L)$ (here
$\Bbb{A}(\L ')=\Bbb{A}^{\#\L '}$, $\Bbb{A}(\L)=\Bbb{A}^{\#\L}$).

\begin{prop}\label{5.22}
$X(\L ')$ is a subvariety of $X(\L)$ if and only if $\L '$ is an embedded sublattice of $\L$.
\end{prop}
\begin{pf}
Under the embedding $X(\L ')\hookrightarrow\Bbb{A}(\L)$, $X(\L ')$ can be identified with
$$\{(x_{\t})_{\t\in\L}\in\Bbb{A}(\L)\mid x_\t=0\text{ if }\t\not\in\L ',\ x_\t x_\f
=x_{\t\v\f}x_{\t\w\f}\text{ for }\t,\f\in\L '\text{ noncomparable }\}.$$
Let $\eta '$ be the generic point of $X(\L ')$. We have $X(\L ')\subset X(\L)$ if and only if
$\eta '\in X(\L)$.

Assume that $\eta '\in X(\L)$. Let $\t$, $\f$ be two noncomparable elements of $\L$ such that
$\t\v\f$, $\t\v\f$ are both in $\L '$. We have to show that $\t,\,\f\in\L '$. If possible, let
$\t\not\in\L '$. This implies $\eta_\t '=0$. Hence either $\eta_{\t\v\f} '=0$, or
$\eta_{\t\w\f} '=0$, since $\eta '\in X(\L)$. But this is not possible (note that $\t\v\f$,
$\t\w\f$ are in $\L '$, and hence $\eta_{\t\v\f} '$ and $\eta_{\t\w\f} '$ are both nonzero).

Assume now that $\L '$ is an embedded sublattice. We have to show that $\eta '\in X(\L)$. Let
$\t$, $\f$ be two noncomparable elements of $\L$. The fact that $\L '$ is a sublattice implies
that if $\eta_{\t\v\f} '$ or $\eta_{\t\w\f} '$ is zero, then either $\eta_\t '$, or
$\eta_\f '$   is  zero. Also, the fact that $\L '$ is an embedded sublattice implies that if
$\eta_\t '$ or $\eta_\f '$ is zero, then either $\eta_{\t\v\f} '$ or $\eta_{\t\w\f} '$ is zero.
Further, when $\t,\,\f,\,\t\v\f,\,\t\w\f\in\L '$,
$$\eta_\t '\eta_\f '=\eta_{\t\v\f} '\eta_{\t\w\f} '.$$
Thus $\eta '$ satisfies the defining equations of $X(\L)$, and hence $\eta '\in X(\L)$.
\end{pf}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%% standard monomial basis

\section{A standard monomial basis for $R(\L)$} \label{s6}
\subsection{}
Let $R(\L)=k[\L]/I(\L)$, $k[\L]$ being as in Section \ref{s3}. Note that $R(\L)$ is a domain (cf.
Theorem \ref{3.7}). Let $X(\L)=\text{Spec}R(\L)$. For $\te\in\L$, let  us consider the
sublattice
$\L_\te=\{\l\in\L\mid\l\le\te\}$ and the ring
$R_\te=R_{\L}/\langle x_\l\mid \l\in\L, \l\not\le\te\rangle\cong R(\L_\te)$
and denote $\text{Spec}R_\te$ by $X(\L_\te)$, or just $X_\te$.  Note that $\L_{\te}$ is an
embedded sublattice (cf. Definition \ref{5.21}). In the sequel, for
$\t\in\L$, we shall denote $\bar{x}_\t$ (in $R(\L)$) by just $x_\t$; similarly,
the restriction $x_\t\bigr|_{X_\te}$ will be also denoted by just $x_\t$.
\subsection{} We take a total order $\succ$ on $\L$ extending the partial order $>$, and
a monomial of degree $r$ in $R(\L)$ will be written as $x_{\t_1}\dots x_{\t_r}$, with
$\t_1\succeq\dots\succeq\t_r$.
\begin{defn}
A monomial  $x_{\t_1}\dots x_{\t_r}$ of degree $r$ is said to be\/ {\em standard on }
$X(\L)$, if $\t_1\ge\dots\ge\t_r$. Such a monomial is said to be\/ {\em standard  on}
$X_\te$ if  $\te\ge\t_1\ge\dots\ge\t_r$.
\end{defn}
\begin{prop}
Standard monomials on $X_\te$ are linearly independent in $R_\te$.
\end{prop}\label{6.4}
\begin{pf}
$R_\te$ being graded, it suffices to prove the linear independence of standard monomials
of a given degree, say $r$. We shall prove this by induction on $l(\te)$ (cf. Definition
\ref{5.5}), and
$r$.  If $l(\te)=0$, or $r=1$, the result is clear. Let then $l(\te)\ge1$, $r>1$, and let
\begin{equation*}
\sum_{i\in I}a_iF_i=0,\qquad a_i\in k^* \tag{$*$}
\end{equation*}
be a linear relation, where $F_i=\t_{i1}\dots\t_{ir}$, $\t_{i1}\ge\dots\ge\t_{ir}$.
If $\t_{i1}<\te$ for some $i$, then restricting $(*)$ to
$X_{\t_{i1}}$, we obtain (by induction hypothesis) that $a_j=0$ for all $j$ with
$\t_{j1}\le\t_{i1}$ (note that $F_j\bigr|_{X_{\t_{i1}}}$ is either identically zero, or it
remains standard on $X_{\t_{i1}}$). Thus we may suppose that $\t_{i_1}=\te$ for all
$i\in I$. Now $x_\te$ can be cancelled out (since $R_\te$ is a domain), and the result follows by
induction on $r$.
\end{pf}
\begin{prop}\label{6.5}
Any monomial in $R_\te$ is standard.
\end{prop}
\begin{pf}
Let $F=x_{\t_1}\dots x_{\t_r}$. If there is an $i$ such that $\t_i\not\ge\t_{i+1}$, this
implies $\t_i$, $\t_{i+1}$ are non-comparable. Then, denoting $\t_i\v\t_{i+1}$,
$\t_i\w\t_{i+1}$ by $\l$, $\m$ respectively, we have
$F=x_{\t_1}\dots x_\l x_\m \dots x_{\t_r}$, and the new expression of $F$ is
lexicographically greater than the previous one, since $\l>\t_i$. Continuing this process,
after a finite number of steps we arrive at a standard monomial expression of $F$
(since $\L$ is finite).
\end{pf}
Combining Propositions \ref{6.4} and \ref{6.5}, we obtain
\begin{thm}\label{6.6}
Standard monomials on $X_\te$ form a basis of $R_\te$.
\end{thm}


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%% CM

\section{Cohen-Macaulayness of $R(\L)$}\label{s7}
In this section we prove the Cohen-Macaulayness of $R(\L)$ using deformation techniques.
Let $S(\L)$ denote the Stanley-Reisner of $\L$, namely
$k[\L]/\langle x_\a x_\b\mid (\a,\b)\in Q\rangle$ (recall that
$Q=\{(\a,\,\b)\mid\a,\,\b\in\L\text{ noncomparable }\}$).

Recall the following (cf. \cite{BGS}):
\begin{thm}\label{7.1}
The ring $S(\L)$ is Cohen-Macaulay.
\end{thm}
We now construct a flat family over $k[t]$ whose general fiber is $R(\L)$ and special
fiber is $S(\L)$. This construction is done in the same spirit as in \cite{GLdef}.
\subsection{}
We first assign positive integers $n_\t$, $\t\in\L$, in such a way that if $\t>\t'$,
then $n_\t>n_{\t'}$ (for example we may take $n_\t=l(\t)$). We choose an integer $N$,
and set $N_\t=N^{l(\t)}$. Then, since $\L$ is finite, we can choose $N$ sufficiently large
so that for any diamond $D(\t,\f,\t\v\f,\t\w\f)$,
\begin{equation*}
N_{\t\v\f}+N_{\t\w\f}>N_\t+N_\f.
\end{equation*}
\begin{thm}\label{7.3}
There exists a flat family over $k[t]$ whose special fiber is $S(\L)$ and general fiber
is $R(\L)$.
\end{thm}
\begin{pf}
For $\t,\f\in Q$ , let $f_{\t,\f}$ be the element in $k[\L]$ given by
\begin{equation*}
f_{\t,\f}=x_{\t}x_{\f}-x_{\t\v\f}x_{\t\w\f}.
\end{equation*}
Then $I(\L)=\langle f_{\t,\f}\mid (\t,\f)\in Q\rangle$. Let us denote $R(\L)$, $S(\L)$,
$k[\L]$, by $R$, $S$, $P$ respectively.
Let   $A=k[t]$, and $P_A=A[x_\a,\a\in \L]$. For $(\t,\f)\in Q$, we define
the element $f_{\t,\f,t}$ in $P_A$ as
\begin{equation*}
f_{\t,\f,t}=x_{\t}x_{\f}-x_{\t\v\f}x_{\t\w\f}t^{N_{\t\v\f}+N_{\t\w\f}-N_{\t}-N_{\f}}.
\end{equation*}
(note that $f_{\t,\f,t}$ is well defined, in view of the choice of $N_\a$'s).

Let $\cal{I}$ be the ideal in $P_A$ generated by $\{f_{\t,\f,t}\mid (\t,\f)\in
Q\}$, and $\cal{R}=P_A/\cal{I}$.
\begin{claim}$(a)$ $\cal{R}$ is $k[t]$-free.

$(b)$ $\cal{R}\otimes_{k[t]}k[t,t^{-1}]\simeq R[t,t^{-1}]$.

$(c)$ $\cal{R}\otimes_{k[t]}k[t]/(t)\simeq S$.
\end{claim}
\begin{pf}
We have
\begin{equation*}
\cal{R}\otimes_{k[t]} k[t]/(t)=P_A/\langle \cal{I}+(t)\rangle =S
\end{equation*}
This proves $(c)$.
 Let $B=k[t,t^{-1}]$, and
$P_B=B[x_{\a}, \a\in \L]$. Let $\tilde{I}$(resp. $\tilde{\cal{I}}$) be the ideal
in $P_B$ generated by $\{f_{\t,\f}\mid (\t,\f)\in Q\}$(resp. $\{f_{\t,\f,t}\mid
(\t,\f) \in Q\}$. We have
\begin{equation}\label{iso1}
P_B/\tilde{I}\simeq R[t,t^{-1}]
\end{equation}
\begin{equation}\label{iso2}
 P_B/\tilde{\cal{I}}\simeq \cal{R}\otimes_{k[t]}k[t,t^{-1}]
\end{equation}
 The automorphism
\begin{equation*}
P_B\simeq P_B,\qquad x_{\a}\mapsto t^{N_{\a}}x_{\a}
\end{equation*}
induces  an isomorphism
\begin{equation}\label{iso3}
P_B/\tilde{I}\simeq P_B/\tilde{\cal{I}}
\end{equation}
From (\ref{iso1}), (\ref{iso2}), (\ref{iso3}) we obtain $(b)$.
Finally, it remains to show $(a)$.
Let $X_{\a}=\bar{x}_{\a}$ (in $\cal{R}=
P_A/\cal{I}$), $f_{\a}=t^{N_{\a}}X_{\a}$ and
\begin{equation*}
\cal{M}=\{f_{\a_1}\dots f_{\a_r}\mid \a_1\ge \dots\ge \a_r,\  r\in \Bbb{Z}_+\}.
\end{equation*}
We shall now show that $\cal{M}$ is a $k[t]$-basis for $\cal{R}$.
We first observe  that $\cal{M}$ is a $k[t]$-basis for $\cal{R}$. We first observe that
any monomial $F=f_{\t_1}\dots f_{\t_r}$ is in fact standard. In order to see this, let
$i$ be such that $\t_i\not\ge\t_{i+1}$. Then using the relation
\begin{equation*}
X_{\t_i}X_{\t_{i+1}}=X_{\t_i\v\t_{i+1}}X_{\t_i\w\t_{i+1}}
t^{N_{\t_i\v\t_{i+1}}+N_{\t_i\w\t_{i+1}}-N_{\t_i}-N_{\t_{i+1}}}\quad ,
\end{equation*}
we obtain
$F=f_{\t_1}\dots f_{\t_{i-1}}f_{\t_i\v\t_{i+1}}f_{\t_i\w\t_{i+1}}\dots f_{\t_r}$.
Continuing thus, we find (as in the proof of Proposition \ref{6.5}) that at each step the
expression for $F$ is lexicographically greater than the expression at the previous
step (here we suppose $\t_1\succeq\dots\succeq\t_r$), and thus we arrive at an expression
for $F$ as a standard monomial (since $\L$ is finite).

It remains to prove the linear independence of $\cal{M}$. Since standard monomials form
a basis for $R$ (cf. Theorem \ref{6.6}), we obtain (by base change), that $\cal{M}$ is a
$k[t,t^{-1}]$-basis for $R[t,t^{-1}]$.
Denoting the isomorphism  $P_B/\tilde{\cal{I}}\simeq R[t,t^{-1}]$ by
$\varphi$, we have $\{\varphi^{-1}(f_{\a_1}\dots f_{\a_r})
\mid \a_1\ge \dots\ge \a_r, r\in \Bbb{Z}_+\}$ is a $k[t,t^{-1}]$-basis for
$\cal{R}[t^{-1}]$. For a monomial $\mn =x_{\t_1}\dots x_{\t_r}$ (in $R[t,t^{-1}]$),
we have $\varphi^{-1}(\mn)=t^{-N_{\mn}}X_{\t_1}\dots X_{\t_r}$, where
$N_{\mn}=\sum_{i=1}^r N_{\t_i}$. Hence we obtain $\{f_{\a_1}\dots f_{\a_r}
\mid \a_1\ge\dots\ge \a_r, r\in\Bbb{Z}_+\}$ is a $k[t,t^{-1}]$-basis for
$\cal{R}[t^{-1}]$ (since $t^{-N_{\mn}}$ is a unit in $k[t,t^{-1}]$).
In particular, we obtain that $\{f_{\a_1}\dots f_{\a_r}\mid \a_1\ge\dots \ge \a_r,
r\in \Bbb{Z}_+\}$ is linearly independent over $k[t,t^{-1}]$, and hence over $k[t]$.
\end{pf}
\end{pf}
Combining Theorems \ref{7.1} and \ref{7.3}, we obtain
\begin{thm}\label{7.4}
The ring $R(\L)$ is Cohen-Macaulay.
\end{thm}


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%% I(d,n)

\section{The distributive lattice $I_{d,n}$ and the variety $X_{d,n}$}\label{s8}
Let
$$I_{d,n}=\{\t=(i_1,\dots,i_d)\mid 1\le i_1<\dots<i_d\le n\}.$$
We consider  the partial order $\ge$ on $I_{d,n}$ given by
$$(i_1,\dots,i_d)\ge (j_1,\dots,j_d)\iff i_1\ge j_1,\dots,i_d\ge j_d.$$

For $\t\in I_{d,n}$, we denote the $j$-th entry in $\t$ by $\t(j)$, $1\le j\le d$.

The following result is well-known (see \cite{Hiller} for example):

\begin{prop}
$(I_{d,n},\ge)$ is a distributive lattice.
\end{prop}
For the rest of this section, the lattice $I_{d,n}$ will be denoted by simply $\L$,
and we  use  the notations introduced in Section \ref{varlat}.

In the discussion below, by a\/ {\em segment} we shall mean a set consisting of
consecutive integers.
\begin{lem}\label{8.2}
We have

$(a)$ The element $\t=(i_1,\dots,i_d)$ is join-irreducible if and only if either $\t$
is a segment, or $\t$ consists of two disjoint segments $(\m,\n)$, with $\m$
starting with $1$.

$(b)$ The element $\t=(i_1,\dots,i_d)$ is meet-irreducible if and only if either $\t$
is a segment, or $\t$ consists of two disjoint segments $(\m,\n)$, with $\n$
ending with $n$.

$(c)$ The element $\t=(i_1,\dots,i_d)$ is join-irreducible and meet-irreducible if and only if either $\t$
is a segment, or $\t$ consists of two disjoint segments $(\m,\n)$, with $\m$ starting with
$1$ and $\n$ ending with $n$.
\end{lem}
\begin{pf}
We first observe that $(i_1,\dots,i_d)$ is join-irreducible if and only if
$(n+1-i_d,\dots,n+1-i_1)$ is meet-irreducible. Thus it suffice to prove part $(a)$. It is easily
seen that $\t=(i_1,\dots,i_d)$ is a cover for $(j_1,\dots,j_d)\in\L$ if and only if
$\{j_1,\dots,j_d\}$ is obtained from $(i_1,\dots,i_d)$ by replacing $i_t$ by $i_t-1$ for precisely
one $t$, and this proves $(a)$. Part $(c)$ follows from $(a)$ and $(b)$.
\end{pf}

\subsection{}
For a join-irreducible or meet-irreducible element $\t\in\L$ we say that $\t$ is of
Type I (resp. Type II) if $\t$ consists of just one segment (resp. two disjoint segments), as
in the description given by Lemma \ref{8.2}. We denote by $J_{\L}^{(I)}$, $JM_{\L}^{(I)}$ (resp.
$J_{\L}^{(II)}$, $JM_{\L}^{(II)}$) respectively the set of  elements of $J_{\L}$, $JM_{\L}$ of
Type I (resp. Type II). Note that
$$JM_{\L}^{(I)}=\{(i+1,\dots,i+d)\mid 0\le i\le n-d\},$$
and
$$JM_{\L}^{(II)}=\{(1,\dots,j,n+j+1-d,\dots,n)\mid 1\le j\le d-1\}.$$
For $\t_i=(i+1,\dots,i+d)\in JM_{\L}^{(I)}$ and  $\f_j=((1,\dots,j,n+j+1-d,\dots,n)\in
JM_{\L}^{(II)}$, let $\l_{ij}=\t_i\v\f_j$, $\m_{ij}=\t_i\w\f_j$.  Note that
$\t_0=\widehat{0}$ and $\t_{n-d}=\widehat{1}$. Clearly,
$$\l_{ij}=(i+1,\dots,i+j,n+1+j-d,\dots,n),\quad\m_{ij}=(1,\dots,j,i+j+1,\dots,i+d).$$

In the sequel, $X_{d,n}$ will denote the variety $X(\L)$, for
$\L=I_{d,n}$.

%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%% boundary components


\section{The irreducible components of $X_{d,n}\setminus X_{d,n}^{\circ}$}\label{s9}
We preserve the notations of Section \ref{s8}.  As in Section \ref{s4}, let
$X_{d,n}^{\circ}=\{(x_1,\dots,x_N)\in X_{d,n}\mid x_i\ne 0\text{ for all } i\}$.
\begin{lem}\label{9.1}
Let $\te\in J_{\L}\setminus JM_{\L}$, say $\te =(1,\dots,j,t+1,\dots,t+d-j)$, where $j<t<n+j-d$.
Then
$\te =\m_{ij}$, where $i=t-j$.
\end{lem}
The result follows from the definition of $\m_{ij}$ (note that $0<i<n-d$).

Let $$\L_i^{(r)}=\L\setminus\{\t\in\L\mid\t (r)=i+r\}$$ where $0\le i\le n-d$, and $1\le r\le d$.

\begin{prop}\label{9.3}
With notations as above, $\L_i^{(r)}$ is an embedded sublattice of $\L$.
\end{prop}
\begin{pf}
Let $\g$, $\d\in\L_i^{(r)}$. Then clearly $\g\v\d$, $\g\w\d\in\L_i^{(r)}$. Similarly, if $\g$,
$\d$ in
$\L$ are such $\g\v\d$, $\g\w\d\in\L_i^{(r)}$, again it is clear that $\g$, $\d\in\L_i^{(r)}$.
\end{pf}

\begin{prop}\label{9.4} Let notations be as above. For $0\le i\le n-d$, let
$$Y_i=X_{d,n}\cap\{x_{\t_i}=0\}.$$
 Then

$(1)$ $Y_0$ and $Y_{n-d}$ are irreducible

$(2)$ the irreducible components of $Y_i$, $1\le i\le n-d-1$, are precisely $X(\L_i^{(r)})$,
$1\le r\le d$.
\end{prop}
\begin{pf}
In order to prove $(1)$, it is enough to observe that $Y_0=X(\L\setminus\{\widehat{0}\})$,
$Y_{n-d}=X(\L\setminus\{\widehat{1}\})$. Now we prove $(2)$. We have, in view
of Propositions \ref{5.22} and \ref{9.3}, that $X(\L_i^{(r)})$,
$1\le r\le d$ is an irreducible subvariety of $X(\L)$. Also $X(\L_i^{(r)})\subset Y_i$ (since
$\t_i\not\in\L_i^{(r)}$).

Let now $Z\subset Y_i$, $Z$ irreducible. Let $\eta$ be the generic point of $Z$. Let $\t\in\L$,
$\t\ne\t_i$ such that $\eta_\t=0$ (note that, since $\eta_{\t_i}=0$, for at least one $\t\in\L$,
$\t\ne\t_i$, $\eta_\t =0$).
\begin{claim}
$\t\in\L_i^{(r)}$, for some $r$, $1\le r\le d$.
\end{claim}
\begin{pf}
Assume that $\t\not\in\L_i^{(r)}$, for all $1\le r\le d$. This implies $\t (r)=i+r$ for all $1\le
r\le d$. But this implies that $\t =\t_i$, contradicting the choice of $\t$. Hence our assumption
is wrong, and the Claim follows.
\end{pf}
Now the  Claim implies that $\eta\in X(\L_i^{(r)})$, for some $r$, $1\le r\le d$, and hence
$Z\subset  X(\L_i^{(r)})$.
\end{pf}
\begin{cor}
The subvariety $X(\L_i^{(r)})$, $1\le i\le n-d-1$, $1\le r\le d$,  of $X_{d,n}$, has codimension
$1$.
\end{cor}
\subsection{} We next determine the irreducible components of $X_{d,n}\cap\{x_\f=0\}$, for $\f\in
JM_{\L}$
of Type II. Let
$$\f =\f_j =(1,\dots,j,n+1+j-d,\dots,n),$$ where $1\le j\le d-1$. Fix $r$, with $j+1\le r\le
n+j-d$, and let
$$
\begin{aligned}
\l_j^{(r)} =(r-j,r-j+1,\dots,&r-1,n+j+1-d,n+j+2-d,\dots,n)\\
\m_j^{(r)} =(1,\dots,j,r+1,&r+2,\dots,r+d-j),
\end{aligned}
$$
where $\l_j^{(r)}(j)=r-1$.
Note that
$$\te\in [\m_j^{(r)},\l_j^{(r)}]\iff\te (j+1)\ge r+1,\text{ and }\te (j)\le r-1$$
(and hence
$\te\not\in [\m_j^{(r)},\l_j^{(r)}]\iff\te (j+1)\le r,\text{ or }\te (j)\ge r$). Let
$$\L_j^{(r)}=\L\setminus [\m_j^{(r)},\l_j^{(r)}],$$
 where $1\le j\le d-1$, $j+1\le r\le n+j-d$.

\begin{prop}\label{9.7}
With notations as above, $\L_j^{(r)}$ is an embedded sublattice of $\L$.
\end{prop}
\begin{pf}
Let $\g$, $\d\in\L_j^{(r)}$. Then either $\g (j)\ge r$, or $\g (j+1)\le r$, and $\d (j)\ge r$, or
$\d (j+1)\le r$.
\begin{claim}
$\g\v\d$, $\g\w\d\in\L_j^{(r)}$.
\end{claim}
\begin{pf}
We distinguish four cases.

\noindent {\it Case 1}\,:\quad $\g (j)\ge r$, $\d (j)\ge r$

\noindent This implies that $(\g\v\d) (j)\ge r$, $(\g\w\d) (j)\ge r$, and hence
$\g\v\d,\,\g\w\d\in\L_j^{(r)}$.

\noindent {\it Case 2\,}:\quad $\g (j)\ge r$, $\d (j+1)\le r$

\noindent This implies that $(\g\v\d) (j)\ge r$, $(\g\w\d) (j+1)\le r$, and hence
$\g\v\d,\,\g\w\d\in\L_j^{(r)}$.

\noindent {\it Case 3\,}:\quad $\g (j+1)\le r$, $\d (j)\ge r$

\noindent This is similar to the previous case.

\noindent {\it Case 4\,}:\quad $\g (j+1)\le r$, $\d (j+1)\le r$

\noindent In this case, $(\g\v\d) (j+1)\le r$, $(\g\w\d) (j+1)\le r$, and hence
$\g\v\d,\,\g\w\d\in\L_j^{(r)}$.
\end{pf}

\begin{claim}
Let $\g$, $\d$ be two noncomparable elements of $\L$ such that $\g\v\d$, $\g\w\d\in\L_j^{(r)}$.
Then $\g$, $\d\in\L_j^{(r)}$.
\end{claim}
\begin{pf}
Here, we distinguish three cases.

\noindent {\it Case 1\,}:\quad $(\g\w\d)(j)\ge r$

\noindent This implies $\g (j)\ge r$, $\d (j)\ge r$. Hence $\g$, $\d\in\L_j^{(r)}$.

\noindent {\it Case 2\,}:\quad $(\g\v\d)(j+1)\le r$

\noindent This implies $\g (j+1)\le r$, $\d (j+1)\le r$. Hence $\g$, $\d\in\L_j^{(r)}$.

\noindent {\it Case 3\,}:\quad $(\g\v\d)(j)\ge r$, $(\g\w\d)(j+1)\le r$

Now $(\g\v\d)(j)\ge r$ implies that at least one of $\{\g (j),\,\d (j)\}$ is $\ge r$.
Assume that $\g (j)\ge r$. This implies $\g (j+1)\ge r+1$, and hence $\d (j+1)\le r$
(since
$(\g\w\d)(j+1)\le r$). Thus, $\g (j)\ge r$, and $\d (j+1)\le r$. Hence $\g$, $\d\in\L_j^{(r)}$.
\end{pf}
The above two claims show that $\L_j^{(r)}$ is an embedded sublattice.
\end{pf}

\begin{prop}\label{9.8}
Let notations be as  above. For $1\le j\le d-1$, let $$Y_j=X_{d,n}\cap\{x_{\f_j}=0\}.$$ Then the
irreducible components of $Y_j$ are precisely $X(\L_j^{(r)})$, $j+1\le r\le n+j-d$.
\end{prop}
\begin{pf}
We have, in view of Propositions \ref{5.22} and \ref{9.7}, that $X(\L_j^{(r)})$ is an irreducible
subvariety of
$X(\L)$.   Also $X(\L_j^{(r)})\subset Y_j$ (since $\f_j\not\in \L_j^{(r)}$). Let now
$Z\subset Y_j$,
$Z$ irreducible. Let $\eta$ be the generic point of $Z$. Consider an element $\t\in\L$,
$\t\ne\f_j$ such that $\eta_\t=0$ (note that, since $\eta_{\f_j}=0$, for at least one $\t\in\L$,
$\t\ne\f_j$, $\eta_{\t}=0$).
\begin{claim}
$\t\in\L_j^{(r)}$, for some $j+1\le r\le n+j-d$.

\end{claim}
\begin{pf}
Assume that $\t\not\in\L_j^{(r)}$, for all $j+1\le r\le r+j-d$. Then for $r=j+1$, we have
$\t(j)<j+1$. Hence we obtain $\t(j)=j$. Similarly, for $r=n+j-d$, we have $\t(j+1)>n+j-d$. Hence
$\t(j+1)=n+1+j-d$ (note that $\t(j+1)\le n+1+j-d$ for any $\t\in\L$). Thus we obtain that
$\t=(1,\dots,j,n+1+j-d,\dots,n)$, i.e. $\t=\f_j$, which is not true, by our choice of $\t$. Hence
our assumption is wrong, and the Claim follows
\end{pf} Now, Claim implies that $\eta\in X(\L_j^{(r)})$, for some $r$, with $j+1\le r\le
n+j-d$, and hence $Z\subset X(\L_j^{(r)})$, for some $r$, with $j+1\le r\le n+j-d$.
\end{pf}
\begin{cor}
The subvariety $X(\L_j^{(r)})$, $1\le j\le d-1$, $j+1\le r\le n+j-d$, of $X_{d,n}$, has
codimension $1$.
\end{cor}
\begin{prop}\label{9.9}
The irreducible components of $X_{d,n}\setminus X_{d,n}^{\circ}$ are precisely the subvarieties
$X(\L_i^{(r)})$, $1\le i\le n-d-1$, $1\le r\le d$,  $X(\L_j^{(r)})$, $1\le j\le d-1$, $j+1\le
r\le n+j-d$, and $X(\L\setminus\{\widehat{0}\})$, $X(\L\setminus\{\widehat{1}\})$.
\end{prop}
\begin{pf}
Let $Z\subset X_{d,n}\setminus X_{d,n}^{\circ}$, $Z$ irreducible, and let $\eta$ be the generic
point of $Z$. In view of Propositions \ref{9.4} and \ref{9.8}, it is enough to show that there
exists a $\t\in JM_{\L}$ such that $\eta_\t=0$.

Assume that $\eta_\t\ne 0$, for any $\t\in JM_{\L}$. Let $\te\in
J_{\L}\setminus JM_{\L}$. Then
$\te$ is a Type I join-irreducible element of $\L$. Let
$$\te=(1,\dots,j,t+1,\dots,t+d-j).$$ We have $\te=\m_{ij}$, where $i=t-j$ (cf. Lemma \ref{9.1}).
Thus
$\te=\t_i\w\f_j$, and hence our assumption implies that $\eta_{\te}\ne 0$. This, together with
our assumption, implies that $\eta_\d\ne 0$, for any $\d\in\J_{\L}$. This implies $\eta\in
X_{d,n}^{\circ}$ (cf. Lemma \ref{5.20}), which is not true. Hence our assumption is wrong.
\end{pf}
\begin{lem}\label{merged}
We have

$(a)$ for $1\le i\le n-d-1$, $JM_{\L}\setminus\{\t_i\}\subset\L_i^{(r)}$, where
$\t_i=(i+1,\dots,i+d)$,

$(b)$ for $1\le j\le d-1$, $JM_{\L}\setminus\{\f_j\}\subset\L_j^{(r)}$, where
$\f_j=(1,\dots,j,n+j+1-d,\dots,i+d)$.
\end{lem}
\begin{pf}
$(a)$ Let $\te\in JM_{\L}$.
First, let $\te$ be of Type I, say $\te = (j+1,\dots,j+d)$, where $j\ne i$. Then, clearly,
$\te\in\L_i^{(r)}$. Now let $\te$ be of Type II, say $\te =(1,\dots,j,n+1+j-d,\dots,n)$. Then
$\te (r)=r$ if $r\le j$, and $\te (r)=n-d+r$ if $r>j$. In either case, it is clear that
$\te\in\L_i^{(r)}$

$(b)$ Let $\te\in JM_{\L}$. First, let  $\te$ be of Type I, say $\te = (i+1,i+2,\dots,i+d)$. We
have $\te (j)=i+j$, $\te (j+1)=i+j+1$. Hence $\te (j)\ge r$, if $i+j\ge r$, and $\te (j+1)\le r$,
if $i+j+1\le r$. Thus $\te\in\L_j^{(r)}$.

Now let $\te$ be of Type II, say $\te =(1,\dots,s,n+1+s-d,\dots,n)$. If $j<s$, then $\te
(j+1)=j+1\le r$, and hence
$\te\in\L_j^{(r)}$. If $j\ge s$, then $\te (j)=n-d+j$, and $\te (j+1) =n-d+j+1$. Hence either
$n-d+j+1\le r$, or $n-d+j\ge r$. Thus $\te\in\L_j^{(r)}$.
\end{pf}
\begin{prop}\label{9.10}
Let $Y$ be an irreducible  component of $X_{d,n}\setminus X_{d,n}^{\circ}$. Let $P$ be the generic
point of
$Y$. Then
$P_\t=0$, for precisely one element $\t\in JM_{\L}$.
\end{prop}
\begin{pf}
If $Y$ is equal to  $X(\L\setminus\{\widehat{0}\})$ or $X(\L\setminus\{\widehat{1}\})$ (cf.
Proposition \ref{9.9}), the assertion is obvious. For the other components the result follows from
Proposition \ref{9.9}, and Lemma \ref{merged}.
\end{pf}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%% normality

\section{Normality of $X_{d,n}$}\label{s10}
In this section we prove that the variety $X_{d,n}$ is normal. We already know that $X_{d,n}$
is Cohen-Macaulay (cf. Section \ref{s7}), and in view of Serre Criterion, it suffices to show that
$X_{d,n}$ is nonsingular in codimension $1$. In view of Proposition \ref{9.9}, it is enough to
prove that  generic points of $X(\L\setminus\{\widehat{0}\})$, $X(\L\setminus\{\widehat{1}\})$,
$X(\L_i^{(r)})$, $1\le i\le n-d-1$, $1\le r\le d$, $X(\L_j^{(r)})$, $1\le j\le d-1$, $j+1\le r\le
n+j-d$ are smooth points. We prove this using the Jacobian Criterion.


Let $\J$ be the Jacobian matrix of $X_{d,n}$, where the rows of $\J$ are indexed by
$$\{f_{\t,\f}=x_\t x_\f-x_{\t\v\f}x_{\t\w\f}\mid \t,\f\in\L \text{ noncomparable }\},$$ and the
columns are indexed by $\{x_\te\mid\te\in\L\}$. For simplicity, sometimes we consider the rows
being indexed by the diamonds in $\L$, and the columns being indexed by just the elements of $\L$.

Recall that for any $P\in X_{d,n}$, if $\J_P$ is the Jacobian matrix evaluated at $P$, then
$\text{rank\,}\J_P\le \text{codim\,}_{\Bbb{A}(\L)}X_{d,n}$, with equality if and only if $P$ is
a smooth point.
\begin{prop}\label{10.1}
Let $P\in X_{d,n}$ be such that $P_\t\ne 0$ for all $\t\in
JM_{\L}^{(I)}\setminus\{\t_0,\t_{n-d}\}$. Then $P$ is a smooth point.
\end{prop}
\begin{pf}
Let $\J_P$ be the Jacobian matrix evaluated at $P$. We shall now exhibit a submatrix of $\J_P$
of maximal rank, equal to
$$\text{codim\,}_{\Bbb{A}(\L)}X_{d,n}=\binom{n}{d}-d(n-d)-1.$$
For each $i$, $2\le i\le n-d$ , let
$$Z_i=\{\te\in\L\mid\te(d)=i+d\}\setminus\{\t_{ij},\  0\le j\le d-1\},$$
where for $0\le j\le d-1$,
$$\t_{ij}=(i,i+1,\dots,i+j-1,i+j+1,i+j+2,\dots,i+d)$$
(note that $\t_{i-1}<\t_{id-1}<\dots<\t_{i0}=\t_i$, where $\t_{i-1}=(i,i+1,\dots,i+d-1)$ and
$\t_i=(i+1,i+2,\dots,i+d)$, as defined in Section \ref{s8}).

Now, $\t_{ij}$, $0\le j\le d-1$ are precisely the elements with the properties $\t_{ij}(d)=i+d$,
and $\t_{ij}>\t_{i-1}$. Hence $\t_{i-1}$ is not comparable with any $\te$ in $Z_i$. Now,
for each $\te\in Z_i$, we associate the row indexed by $f_{\t_{i-1},\te}$. We consider the
submatrix $\J'$ of $\J_P$ with columns indexed by $\{\te\in Z_i\mid 2\le i\le n-d\}$, and rows
indexed by $\{f_{\t_{i-1},\te}\mid \te\in Z_i,\ 2\le i\le n-d\}$. Then $\J'$ is a square matrix,
of size equal to $\sum_{i=2}^{n-d}\# Z_i$ (note that the $Z_i$'s are disjoint). Now the set
$\{\te\in\L\mid\te(d)=i+d\}$ is in bijection with $I_{d-1,\{1,\dots,i+d-1\}}$ (here, for $r<s$,
$I_{r,\{n_1,\dots,n_s\}}$ denotes the set of all $r$-tuples $\ui=(i_1,\dots,i_r)$, where
$i_1<\dots<i_r$, and $\{i_1,\dots,i_r\}\subset\{n_1,\dots,n_s\}$). Hence
$$\# Z_i=\binom{i+d-1}{d-1}-d,$$ and
$$
\begin{aligned}
\#\bigcup_{i=2}^{n-d}Z_i&=\sum_{i=2}^{n-d}\bigg[\binom{i+d-1}{d-1}-d\bigg]=\binom{n}{d}-
\binom{d}{d-1}-1-d(n-d-1)\\
&=\binom{n}{d}-d(n-d)-1
\end{aligned}
$$
(note that $\sum_{j=1}^{m-r+1}\binom{m-j}{r-1}=\binom{m}{r}$).

Now, a typical row in $\J'$ is indexed by
$$f_{\t_{i-1},\te}=x_{\t_{i-1}}x_{\te}-x_{\t_{i-1}\v\te}x_{\t_{i-1}\w\te},$$
where $\te\in Z_i$. The only variable appearing in $f_{\t_{i-1},\te}$, which is an index for a
column of $\J'$, is $x_\te$. Further, the entry in $\J'$ in the row indexed by
$f_{\t_{i-1},\te}$ and column indexed by $x_\te$ is nonzero (by hypothesis). Thus, $\J'$ is a
diagonal matrix, with nonzero diagonal entries. Henc $\J'$ has rank equal to its size, which is
$$\binom{n}{d}-d(n-d)-1=\text{codim\,}_{\Bbb{A}(\L)}X_{d,n}.$$
Thus we obtain that $P$ is a smooth point of $X_{d,n}$.
\end{pf}
\begin{prop}\label{10.2}
Let $P\in X_{d,n}$ be such that $P_\t\ne 0$, for all $\t\in JM_{\L}^{(II)}$. Then $P$ is a
smooth point of $X_{d,n}$.
\end{prop}
\begin{pf}
Let $\J_P$ be the Jacobian matrix evaluated at $P$. We shall now exhibit a submatrix of $\J_P$
of maximal rank. Fix $r$ and $i$, $1\le r\le d-1$, $r+1\le i\le n-d+r-1$. Set
$$Z_r^{(i)}=\{\te\in\L\mid\te(r)=i,\text{ and }\te(t)=t,\text{ for }t<r\}\setminus\{\te_i\},$$
where $\te_i=(1,\dots,r-1,i,n+r-d+1,\dots,n)$.
Let
$$Z_r=\bigcup_{i=r+1}^{n-d+r-1}Z_r^{(i)}.$$
Then $$\bigcup_{s\le r}Z_s=\{\t\in\L\mid\t\text{ and }\f_r\text{ are noncomparable }\},$$
where $\f_r=(1,\dots,r,n+r-d+1,\dots,n)$, $1\le r\le d-1$, as defined in Section \ref{s8}.
Let $\J'$ be the submatrix of $\J_P$ with columns indexed by $\{\te\in Z_r\mid 1\le r\le
d-1\}$, and rows indexed by $\{f_{\f_r,\te}\mid\te\in Z_r,1\le r\le d-1\}$. Then $\J'$ is a
square matrix of size equal to $\sum_{r=1}^{d-1}\# Z_r$ (note that the $Z_r$'s are disjoint).
Now, $\{\te\in\L\mid\te(r)=i,\text{ and }\te(t)=t,\text{ for }t<r\}$ is in bijection with
$I_{d-r,\{i+1,\dots,n\}}$. Hence
$$\# Z_r^{(i)}=\binom{n-i}{d-r}-1,$$
and
$$
\begin{aligned}
\# Z_r&=\sum_{i=r+1}^{n-d+r-1}\bigg[\binom{n-i}{d-r}-1\bigg]=\binom{n-r}{d-r+1}-1-(n-d-1)\\
&=\binom{n-r}{d-r+1}-(n-d)=\binom{n-r}{n-d-1}-(n-d).
\end{aligned}
$$
Hence the size of $\J'$ is equal to
$$
\begin{aligned}
&\sum_{r=1}^{d-1}\bigg[\binom{n-r}{n-d-1}-(n-d)\bigg]=\binom{n}{n-d}-
\binom{n-d}{n-d-1}-1-(d-1)(n-d)\\
&=\binom{n}{d}-(n-d)-1-(d-1)(n-d)=\binom{n}{d}-d(n-d)-1.
\end{aligned}
$$
As in the proof of Proposition \ref{10.1}, we find that $\J'$ is a diagonal matrix, with nonzero
diagonal entries. Hence the  rank of $\J'$ is equal to its size, which is
$$\binom{n}{d}-d(n-d)-1=\text{codim\,}_{\Bbb{A}(\L)}X_{d,n}.$$
Thus $P$ is a smooth point of $X_{d,n}$.
\end{pf}
\begin{prop}\label{10.3}
The variety $X_{d,n}$ is nonsingular in codimension $1$.
\end{prop}
\begin{pf}
Let $Y$ be an irreducible component of $X_{d,n}\setminus X_{d,n}^{\circ}$. Let $P$ be the generic
point of
$Y$. Then, by Proposition \ref{9.10},
$P_\t=0$ for precisely one element $\t\in JM_{\L}$. Hence the result follows from
Propositions \ref{10.1} and \ref{10.2}.
\end{pf}
Proposition \ref{10.3} and Theorem \ref{7.4} yield the following
\begin{thm}\label{10.4}
The variety $X_{d,n}$ is a normal toric variety.
\end{thm}
\subsection{}
Let $A_{d,n}$ denote the homogeneous coordinate ring of $G_{d,n}$ (the Grassmannian of
$d$-planes in $k^n$) for the Pl\" ucker embedding. We recall (cf. \cite{GLdef}):
\begin{thm}\label{10.6}
There exists a flat family whose general fiber is $A_{d,n}$, and whose special fiber is
$R(I_{d,n})$.
\end{thm}
\begin{rem}
The above result is also proved in \cite{St} using SAGBI theory.
\end{rem}
Combining Theorems \ref{10.4} and \ref{10.6}, we obtain
\begin{thm}\label{10.8}
The Grassmannian $G_{d,n}$ degenerates to the normal toric variety $X_{d,n}$.
\end{thm}


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%% singular locus conjecture

\section{A conjecture on the singular locus of $X_{d,n}$}\label{s11}
In this section we prove a partial result towards the determination of the singular locus of
$X_{d,n}$. We also state a conjecture regarding the description of the singular locus of
$X_{d,n}$.

We assume that $n\ge 4$ (note that for $n\le 3$, $X_{d,n}$ is smooth). Let notations be as in
Section \ref{s8}. Let $\L_{ij}=\L\setminus[\m_{ij},\l_{ij}]$, $0\le i\le n-d$, $1\le j\le d-1$.

\begin{lem} Let $i=0$, and $1\le j\le d-1$. Then $\L_{0j}$ is an embedded sublattice. Further,
$JM^{(I)}\setminus\{\t_0\}\subset \L_{0j}$.
\end{lem}
\begin{pf}
We have $\t_0=(1,\dots,d)$, $\f_j=(1,\dots,j,n+1+j-d,\dots,n)$, and hence $\m_{0j}=\t_0$, and
$\l_{ij}=\f_j$. Hence $\te\in [\t_0,\f_j]$ if and only if $\te(t)=t$, for $1\le t\le j$. Therefore
$\te\in\L_{0j}$ if and only if $\te(t)\ne t$, for some $t$, $1\le t\le j$. Let $\g,\, \d$ be
two noncomparable elements in $\L_{0j}$. Let $t_1,\, t_2\le j$ be such that $\g(t_1)\ne t_1$,
$\d(t_2)\ne t_2$. Then, letting $t=\max (t_1,t_2)$, we have $\g\v\d (t)\ne t$, $\g\w\d (t)\ne t$.
Thus $\g\v\d,\, \g\w\d\in\L_{0j}$.

Let now $\g,\,\d$ be two noncomparable elements of $\L$ such that $\g\v\d,\,\g\w\d\in\L_{0j}$. Let
$t\le j$ be such that $\g\w\d (t)\ne t$. This implies $\g\v\d (t)\ne t$. Then, clearly $\g
(t)\ne t$, $\d (t)\ne t$. Thus $\g,\,\d\in\L_{0j}$. Thus we obtain that $\L_{0j}$ is an embedded
sublattice. Let $\t\in JM^{(I)}\setminus\{\t_0\}$, say $\t=(i+1,\dots,i+d)$, where $i\ge 1$.
Then clearly $\t\in\L_{0j}$ (note for example that $\t(1)\ne 1$).
\end{pf}
\begin{cor}
Let $P$ be the generic point of $X(\L_{0j})$. Then $P$ is a smooth point of $\L$.
\end{cor}
\begin{pf}
We have $P_{\t}\ne 0$, for $\t\in JM^{(I)}\setminus\{\t_0\}$. Hence the result follows from
Proposition 10.1.
\end{pf}
\begin{lem}
Let $i=n-d$, and $1\le j\le d-1$. Then $\L_{n-d\, j}$ is an embedded sublattice. Further,
$JM^{(I)}\setminus\{\t_{n-d}\}\subset \L_{n-d\, j}$.
\end{lem}
\begin{cor}
Let $P$ be the generic point of $X(\L_{n-d\, j})$. Then $P$ is a smooth point of $\L$.
\end{cor}

The proofs of Lemma 11.3, and Corollary 11.4 are similar to those of Lemma 11.1, and Corollary
11.2 respectively.

\begin{lem}\label{11.1}
Let $1\le i\le n-d-1$, $1\le j\le d-1$. Then $\L_{ij}$ is an embedded sublattice, and
$$\text{codim\,}_{\Bbb{A}(\L_{ij})}X(\L_{ij})<\text{codim\,}_{\Bbb{A}(\L)}X(\L).$$
\end{lem}
\begin{pf}
For $\te\in\L$ we have
$$\te\in[\m_{ij},\l_{ij}]\iff\te (j)\le i+j,\text{ and }\te (j+1)\ge i+j+1$$
\noindent (here, by $\te(j)$, we mean the $j$-th entry in the $d$-tuple $\te$).
Let $\g$, $\d$ two noncomparable elements in $\L_{ij}$. We have either $\g (j)\ge i+j+1$, or $\g
(j+1)\le i+j$. Similarly, we have either $\d (j)\ge i+j+1$, or $\d (j+1)\le i+j$.
\begin{claim}
$\g\v\d,\,\g\w\d\in\L_{ij}$.
\end{claim}
The proof is the same as in the proof of the first Claim in Proposition \ref{9.7}.
\begin{claim}
Let $\g$, $\d$ be two noncomparable  elements of $\L$ such that $\g\v\d$, $\g\w\d\in\L_{ij}$.
Then $\g$, $\d\in\L_{ij}$.
\end{claim}
The proof is the same as in the proof of the second Claim in Proposition \ref{9.7}.

The above two claims show that $\L_{ij}$ is an embedded sublattice. It remains to prove the
inequality
$$\text{codim\,}_{\Bbb{A}(\L_{ij})}X(\L_{ij})<\text{codim\,}_{\Bbb{A}(\L)}X(\L).$$

If $\te=(k+1,\dots,k+d)\in J_{\L}^{(I)}$, $k\ne i$, is a join-irreducible element of
Type I distinct from $\t_i$, then $\te (j)=k+j\ge i+j+1$, if $k\ge i+1$, and $\te (j+1) =k+j+1\le
i+j$, if $k\le i-1$. Further, the unique element $\te'$ of $\L$ such that $(\te,\,\te')$ is a
cover in $\L$ is given by $\te'=(k,k+2,\dots,k+d)$. It is easily seen that if either $k\ne i+1$,
or $j\ne 1$ then $\te'\in\L_{ij}$, and is the unique element of $\L_{ij}$ such that
$(\te,\,\te')$ is a cover in $\L_{ij}$. Let then $k=i+1$, and $j=1$. In this case, we observe
that if
$\d=(i,i+1,i+4,\dots,i+d+1)$, then $\d\in\L_{ij}$, and $\d$ is the unique element of $\L_{ij}$
such that $(\te,\d)$ is a cover in $\L_{ij}$ (note that $[\d,\,\te]\cap\L_{ij}=\{\d,\,\te\}$).
 Thus $\te\in J_{\L_{ij}}$.

 Let now  $\te =(1,\dots,s,t+1,\dots,t+d-s)\in
J_{\L}^{(II)}$, where $t\ne n+j-d$, if $s=j$, be a join-irreducible element of Type II
distinct from $\f_j$. We distinguish three cases.

\noindent {\it Case 1\,}:\quad  $s<j$

\noindent We have $\te (j)=t+j-s$, $\te (j+1)=t+j+1-s$. Hence, if $t-s\ge i+1$, then $\te (j)\ge
i+j+1$, and $\te\in \L_{ij}$. If $t-s\le i-1$, then $\te (j+1)\le i+j$, and $\te\in
\L_{ij}$.  Further, the unique element $\te'$ of $\L$ such that $(\te,\,\te')$ is a
cover in $\L$ is given by $\te'=(1,\dots,s,t,t+2,\dots,t+d)$. It is easily seen that
$\te'\in\L_{ij}$, and is the unique element of $\L_{ij}$ such that $(\te,\te')$ is a cover, except
when $s+1=j$, and
$t-s=i+1$. Thus $\te\in J_{\L_{ij}}$, except when $s+1=j$, and $t-s=i+1$. In this case, we observe
that if
$\d=(1,\dots,s,i+s,i+s+1,t+3,\dots,t+d-j)$, then $\d\in\L_{ij}$, and is the unique element of
$\L_{ij}$ such that $(\te,\,\d)$ is a cover in $\L_{ij}$. Thus $\te\in J_{\L_{ij}}$.

\noindent {\it Case 2\,}:\quad $s=j$

\noindent We have $\te (j)=j$, $\te (j+1)=t+1$. Hence, if $\te (j+1)\le i+j$, i.e. $t\le i+j-1$,
then $\te\in \L_{ij}$. Further, the unique element $\te'$ of $\L$ such that $(\te,\,\te')$
is a cover in $\L$ (namely $\te'= (1,\dots,j,t,t+2.\dots,t+d-j)$) also belongs to $\L_{ij}$, and
is the unique element of $\L_{ij}$ such  that $(\te,\,\te')$ is a cover in $\L_{ij}$. Hence
$\te\in J_{\L_{ij}}$.

\noindent {\it Case 3\,}:\quad $s>j$

\noindent We have $s\ge j+1$, and hence $\te (j+1) =j+1\le i+j$. Hence $\te\in \L_{ij}$.
As in \break {\it Case 2}, if $\te'$ is the unique element of $\L$ such that $(\te,\te')$ is a
cover in $\L$, then $\te'\in\L_{ij}$, and is the unique element of $\L_{ij}$ such that
$(\te,\te')$ is a cover in $\L_{ij}$.  Hence $\te\in J_{\L_{ij}}$.

From above, we have
$$\# J_{\L_{ij}}\ge \# J_{\L}-(n-d-i+j+1).$$
On the other hand, we have
$$\# J_{\L}-\# J_{\L_{ij}}=\# [\m_{ij},\l_{ij}]=\binom{i+j}{j}\binom{n-i-j}{d-j}.$$
It can be seen easily (by assuming $d\le [n/2]$, since $I_{d,n}$ is isomorphic to $I_{n-d,n}$)
that $$\binom{i+j}{j}\binom{n-i-j}{d-j}>n-d-i+j+1,$$
i.e.
$$\#\L-\# J_{\L}>\#\L_{ij}-\# J_{\L_{ij}}.$$
The result now follows from this.
\end{pf}

\begin{prop}
For $1\le i\le n-d-1$, $1\le j\le d-1$, we have
 $$X(\L_{ij})\subset\text{Sing\,}(X_{d,n}).$$
\end{prop}
\begin{pf}
Let $P$ be the generic point of $\L_{ij}$. We have $P_\t=0$ for all $\t\in [\m_{ij},\l_{ij}]$.
 Let $\J_P$ be the Jacobian matrix evaluated at $P$. Then
the submatrix of $\J_P$ with rows given by $\{f_{\t,\f}\mid\t,\f\in
[\m_{ij},\l_{ij}],\t,\,\f\text{ noncomparable }\}$, and columns given by $\{x_\t\mid\t\in
[\m_{ij},\l_{ij}]\}$ is the zero matrix. Let $\J'$ be the matrix obtained from $\J_P$ by deleting
these rows and columns. Then we have
\begin{equation*}
\text{rank\,}\J'=\text{rank\,}\J_P.\tag{$1$}
\end{equation*}
Further, $\J'$ is precisely the Jacobian matrix of the variety $X(\L_{ij})\subset
\Bbb{A}_{\L_{ij}}$, evaluated at $P'=(P_\t)_{\t\in\L_{ij}}$. Hence
\begin{equation*}
\text{rank\,}\J'\le\text{codim}_{\Bbb{A}(\L_{ij})}X(\L_{ij})<\text{codim}_{\Bbb{A}(\L)}
X_{d,n}\tag{$2$}
\end{equation*}
(cf. Lemma \ref{11.1}).
The result now follows from ($1$) and ($2$).
\end{pf}
We have the following conjecture on the singular locus of $X_{d,n}$.
\begin{conj}
The irreducible components of $\text{Sing\,}X_{d,n}$ are
precisely $X(\L_{ij})$, $1\le i\le n-d-1$, $1\le j\le d-1$.
\end{conj}



%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Genearlities on $G/Q$}\label{s12}

Let $G$ be a semisimple and simply connected  algebraic group, $T\subset G$ a maximal torus,
and  $B\supset T$, a Borel subgroup. Let $R$ be the root system of $G$ relative
to $T$. Let $R^+$ (resp. $S$) be the system of positive (resp. simple) roots of $R$ with respect
to $B$. Let $Q$ be a parabolic subgroup of $G$ containing $B$. Associated to $Q$, there is a
subset $S_Q$ of $S$ such that $Q$ is the subgroup of $G$ generated by $B$ and
$\{U_{-\a}\mid \a\in R^+_Q\}$, where $R^+_Q=\{\a\in R^+\mid \a=\sum_{\b\in S_Q} a_\b\b\}$ (here,
for $\b\in R$, $U_\b$ denotes the $1$ dimensional unipotent subgroup of $G$). Let
$W_Q$ be the Weyl group of $Q$ (note that $W_Q$ is simply the subgroup of $W$ generated by
$\{s_\a\mid\a\in S_Q\}$.

\subsection{}
In each coset $wW_Q$, there exists a unique element of minimal length (cf. \cite{Bou}). Let
$W_Q^{\text{min}}$ be this set of representatives of $W/W_Q$. The set $W_Q^{\text{min}}$ is
called the {\em set of minimal reperesentatives of} $W/W_Q$. We have
$$W_Q^{\text{min}}=\{w\in W\mid l(ww')=l(w)+l(w'),\text{ for all }w'\in Q\}.$$
The set $W_Q^{\text{min}}$ may be also be characterized as
$$W_Q^{\text{min}}=\{w\in W\mid w(\a)>0, \text{ for all } \a\in S_Q\}$$
(here by a root being $>0$ we mean $\b\in R^+$).
\subsection{}
In each coset $wW_Q$ there exists a unique element of maximal length. Let $W_Q^{\text{max}}$
be the  set of these representatives of $W/W_Q$. We have
$$W_Q^{\text{max}}=\{w\in W\mid w(\a)<0\text{ for all } \a\in S_Q\}.$$
Further, if we denote by $w_Q$ the element of maximal length in $W_Q$, then we have
$$W_Q^{\text{max}}=\{ww_Q\mid w\in W_Q^{\text{min}}\}.$$

In the sequel, given $w\in W$, the minimal (resp. maximal) representative of $wW_Q$  in $W$
will be denoted by $w_Q^{\text{min}}$ (resp. $w_Q^{\text{max}}$).

\subsection{Maximal parabolic subgroups}
The set of maximal parabolic subgroups is in one-to-one correpondence with $S$, namely given
$\a\in S$, the parabolic subgroup $Q$ where $S_Q=S\setminus\{\a\}$ is a maximal parabolic
subgroup, and conversely. We shall denote $Q$, where $S_Q=S\setminus\{\a\}$ by
$P_{\widehat{\a}}$, and refer to it as the {\em maximal parabolic subgroup obtained by
omitting}$\ \a$.

\subsection{Schubert varieties in $G/Q$}\label{12.4}
For $w\in W$, let us denote the coset $wQ$ by $e_{w,Q}$. Then the set of $T$-fixed points in
$G/Q$ for the action given by left multiplication is presisely $\{e_{w,Q}\mid w\in W\}$.
Let $w\in W$, and let $X_Q(w)$ be the Zariski closure of $Be_{w,Q}Q\pmod{Q}$ in $G/Q$. Then
$X_Q(w)$  with the canonical reduced structure is called the Schubert variety in $G/Q$ associated
to $w$. We have the well-known Bruhat decomposition
$$G/Q=\dot{\cup}Be_{w,Q},\qquad X_Q(\te)=\dot{\cup}_{w\le\te}Be_{w,Q},\quad \te\in W.$$

Let us denote by $w_Q^{\text{min}}$ (resp. $w_Q^{\text{max}}$) the minimal (resp. maximal)
representative of $wW_Q$. Let $\pi :G/B\to G/Q$ be the canonical projection. Then it can be
easily seen that
$\pi\bigr|_{X_B(w^{\text{max}})}:X_B(w_Q^{\text{max}})\to X_Q(w)$ is a fibration with fiber
 $\simeq Q/B$ while $\pi\bigr|_{X_B(w^{\text{min}})}:X_B(w_Q^{\text{min}})\to X_Q(w)$ is
birational. In particular, we have $\dim X_Q(w)=\dim X_B(w_Q^{min})$.


\subsection{The big cell and the opposite big cell}\label{12.5}
The $B$-orbit $Be_{w_0}$ in $G/Q$ ($w_0$ being the element of maximal length in $W$) is called
the {\em big cell} in $G/Q$. It is a dense open subset of $G/Q$, and it gets identified with
$R_u(Q)$, the unipotent radical of $Q$, namely the subgroup of $B$ generated by $\{U_\a\mid\a\in
R^+\ \setminus\ R_Q^+\}$  (cf. \cite{B}). Let
$B^-$ be the Borel subgroup of $G$ opposite to $B$, i.e. the subgroup of $G$ generated by $T$ and
$\{U_{-\a}\mid\a\in R^+\}$. The $B^-$-orbit $B^-e_{\text{id},Q}$ is called the {\em opposite big
cell} in $G/Q$. This is again a dense open subset of $G/Q$, and it gets identified with the
unipotent subgroup of $B^-$  generated by $\{U_{-\a}\mid \a\in R^-\ \setminus\ R_Q^-\}$. Observe
that both the big cell and the  opposite big cell can be identified with $\Bbb{A}^{N_Q}$, where
$N_Q=\#\{R^+\ \setminus\  R_Q^+\}$.

For a Schubert variety $X(w)\subset G/Q$, $B^-e_{\text{id}}\cap X(w)$ is called the {\em opposite
 cell} in $X(w)$ (by abuse of language). In general, it is not a cell (except for $w=w_0$). It
is a nonempty affine open subvariety of $X(w)$, and a closed subvariety of the affine space
$B^-e_{\text{id}}$.

\subsection{Equations defining  a Schubert variety}\label{12.6}
Let $L$ be an ample line bundle on $G/Q$. Consider the projective embedding
$G/Q\hookrightarrow \text{Proj}(H^0(G/Q,L))$.
We recall (cf. \cite{RR}) that the homogeneous ideal of $G/Q$ for this embedding is generated
in degree $2$, and any Schubert variety $X$ in $G/Q$ is scheme theoretically
(even at the cone level) the intersection of $G/Q$ with all the hyperplanes in
$\text{Proj}(H^0(G/Q,L))$ containing $X$.

For a maximal parabolic subgroup $P_i$, let us denote the ample generator of
$\text{Pic\,}(G/P_i)$ ($\simeq\Bbb{Z}$) by $L_i$.

Given a parabolic subgroup $Q$, let us denote $S\setminus S_Q$ by $\{\a_1,\dots,\a_t\}$, for
some $t$. Let
\begin{align}
R&=\bigoplus_{\underline{a}}H^0(G/Q, \bigotimes_iL_i^{a_i})\notag\\
R_w&=\bigoplus_{\underline{a}}H^0(X_Q(w), \bigotimes_iL_i^{a_i}),\notag
\end{align}
where $\underline{a}=(a_1,\dots,a_t)\in\Bbb{Z}_+^t$. Let us denote $\bigotimes_iL_i^{a_i}$ by
just $L$. We recall (cf.  \cite{KR}) that the natural map
$$\bigoplus \cal{S}^{a_1}(H^0(G/Q, L_1))\otimes\dots\otimes \cal{S}^{a_1}(H^0(G/Q, L_t))
\to R$$
is surjective, and its kernel is generated as an ideal by elements of total degree $2$.
Further, the restriction map $R\to R_w$ is surjective, and its kernel is generated as an ideal
by elements of total degree $1$.

%%%%%%%%%%%%%%%%%%%%%%%%%
%%% SL(n)/B

\section{Opposite cells in Schubert varieties in $SL(n)/B$}\label{s13}
Let $G=SL(n)$, the special lineear group of rank $n-1$. Let $T$ be the maximal torus consisting
of all the diagonal matrices  in $G$,  and $B$ the Borel subgroup consisting of all the  upper
triangular matrices in
$G$. It is well-known that  $W$ can be identified  with  $\cal{S}_n$, the symmetric group on $n$
letters.

Following \cite{Bou}, we denote the simple roots by $\e_i-\e_{i+1}$, $1\le i\le n-1$ (note that
$\e_i-\e_{i+1}$ is the character sending diag$(t_1,\dots,t_n)$ to $t_it_{i+1}^{-1}$).
Then $R=\{\e_i-\e_j\mid 1\le i,j\le n\}$, and the reflection $s_{\e_i-\e_{i+1}}$ may be
identified with the transposition $(i,j)$ in $\cal{S}_n$.
For $w=(a_1\dots a_n)\in\cal{S}_n$, it is easily seen that $w(\e_i-\e_j)=\e_{a_i}-\e_{a_j}$.



\subsection{The Chevalley-Bruhat order on $\cal{S}_n$}
For $w_1$, $w_2\in W$, we have
$$X(w_1)\subset X(w_2)\iff \pi_d(X(w_1))\subset \pi_d(X(w_2))\text{ for all } 1\le d\le n-1,$$
where $\pi_d$ is the canonical projection $G/B\to G/P_{\widehat{\a_d}}$. Hence we obtain that for
 $(a_1\dots a_n)$, $(b_1\dots b_n)\in\cal{S}_n$,
$$(a_1\dots a_n)\ge (b_1\dots b_n)\iff (a_1\dots a_d)\uparrow\ge (b_1\dots b_d)\uparrow
\text{ for all } 1\le d\le n-1$$
(here, for  an ordered $d$-tuple $(t_1\dots t_d)$ of distinct integers,
$(t_1\dots t_d)\uparrow$ denotes the ordered $d$-tuple obtained from $\{t_1,\dots, t_d\}$ by
arranging its elements in ascending order).

\subsection{The partially ordered set $I_{a_1,\dots,a_k}$}
Let $Q$ be a parabolic subgroup in $SL(n)$. Let  $1\le a_1<\dots <a_k\le n$, such that
$S_Q=S\setminus\{\a_{a_1},\dots,\a_{a_k}\}$ (we follow \cite{Bou} for indexing the simple roots).
Then $Q=P_{\widehat{\a_{a_1}}}\cap\dots\cap P_{\widehat{\a_{a_k}}}$, and
$W_Q=\cal{S}_{a_1}\times\cal{S}_{a_2-a_1}\times\dots\times\cal{S}_{n-a_k}$.
Let
$$I_{a_1,\dots,a_k}=\{(\ui_1,\dots,\ui_k)\in I_{a_1,n}\times\dots\times I_{a_k,n}\mid
\ui_t\subset \ui_{t+1}\text{ for all } 1\le t\le k-1\}.$$
Then it is easily seen that $W_Q^{\text{min}}$ 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 $I_{a_1,\dots,a_k}$, namely, for $\bold{i}=(\ui_1,\dots,\ui_k)$,
$\bold{j}=(\uj_1,\dots,\uj_k)\in I_{a_1,\dots,a_k}$, $\bold{i}\ge\bold{j}\iff\ui_t\ge\uj_t$
for all  $1\le t\le k$.

\subsection{The minimal (maximal) representatives as  permutations}
Let $w\in W_Q$, and let $\bold{i}=(\ui_1,\dots,\ui_k)$ be the  element in $I_{a_1,\dots,a_k}$
which corresponds to
$w_Q^{\text{min}}$.
As a permutation, the element  $w_Q^{\text{min}}$ is given by $\ui_1$, followed by $\ui_2
\setminus\ui_1$ arranged in ascending order, and so on, ending with
$\{1,\dots,n\}\setminus \ui_k$ arranged in ascending order. Similarly, as a permutation, the element
$w_Q^{\text{max}}$ is given by $\ui_1$ arranged in descending order, followed by
$\ui_2\setminus\ui_1$ arranged in descending order, etc..

\subsection{The opposite big cell in $G/Q$}\label{13.4}
Let $Q=\cap_{t=1}^kP_{\widehat{\a_{a_t}}}$. Let $a=n-a_k$, and $Q$ be the parabolic
subgroup consisting of all the elements of $G$ of the form
$$
\begin{pmatrix}
A_1&\ast &\ast &\cdots&\ast&\ast\\
0 &A_2&\ast &\cdots&\ast &\ast\\
\vdots&\vdots&\vdots&&\vdots&\vdots\\
0 &0 &0 &\cdots&A_k&\ast\\
0 &0 &0 &\cdots&0& A
\end{pmatrix},
$$
where $A_t$ is a matrix of size $a_t\times a_t$, $1\le t\le k$, and $A$ is a matrix of size
$a\times a$, and
$x_{ml}=0$, $m>a_t$, $l\le a_t$, $1\le t\le k$. Denote by $O^-$  the subgroup of $G$ generated by
$\{U_\a\mid \a\in R^-\setminus R_Q^-\}$. Then $O^-$ consists of the elements of $G$ of the form
$$
\begin{pmatrix}
I_1&0 &0 &\cdots&0 &0\\
\ast &I_2&0&\cdots& 0&0\\
\vdots&\vdots&\vdots&&\vdots&\vdots\\
\ast &\ast &\ast &\cdots&I_k&0\\
\ast &\ast &\ast &\cdots&\ast& I_a
\end{pmatrix},
$$
where $I_t$ (resp. $I_a$) is the $a_t\times a_t$ (resp. $a\times a$) identity  matrix, and if
$x_{ml}\ne 0$, with $m\ne l$, then $m>a_t$, $l\le a_t$. 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  $B^-e_{\text{id},Q}$ gets identified with $O^-$.

\subsection{Pl\" ucker coordinates of the Grassmannian}\label{13.5}
 Let
$G_{d,n}$ be the Grassmannian variety, consisting of $d$-dimensional subspaces of an
$n$-dimensional vector space $V$.
 Let us identify $V$ with $k^n$, and  denote the standard basis
of $k^n$ by $\{e_i\mid 1\le i\le n\}$. Consider the Pl\" ucker embedding
$f_d:G_{d,n}\hookrightarrow \Bbb{P}(\w^dV)$, where $\w^dV$ is the $d$-th exterior power of $V$.
For $\ui=(i_1,\dots,i_d)\in I_{d,n}$, let $e_{\ui}=e_{i_1}\w\dots\w e_{i_d}$. Then the set
$\{e_{\ui}\mid \ui\in I_{d,n}\}$ is a basis for $\w^dV$. Let us denote the basis of
$(\w^dV)^\ast$ (the linear dual of $\w^dV$) dual to $\{e_{\ui}\mid \ui\in I_{d,n}\}$ by
$\{p_{\uj}\mid \uj\in I_{d,n}\}$. Then $\{p_{\uj}\mid \uj\in I_{d,n}\}$ gives a system of
coordinates for
$\Bbb{P}(\w^dV)$. These are the so-called {\em Pl\" ucker coordinates}.



\subsection{Schubert varieties in the Grassmannian}
Let $Q=P_{\widehat{\a_d}}$. For simplicity of notation, let us denote $P_{\widehat{\a_d}}$ by
just $P_d$.
We have
$$G_{d,n}\simeq G/P_d.$$

Let  $\ui=(i_1,\dots,i_d)\in I_{d,n}$. Then the $T$-fixed
point $e_{\ui,P_d}$ is simply the $d$-dimensional span of $\{e_{i_1},\dots,e_{i_d}\}$. Thus
$X_{P_d}(\ui)$ is simply the Zariski closure of $B[e_{i_1}\w\dots\w e_{i_d}]$ in $\Bbb{P}(\w^dV)$.

In view of the Bruhat decomposition for $X_{P_d}(\ui)$ (cf. \S\ref{12.4}), we have
$$p_{\uj}\bigr|_{X_{P_d}(\ui)}\ne 0\iff\ui\ge\uj.$$
Now using \S\ref{12.6}, we obtain

\begin{rem}
Given $\t\in I_{d,n}$, $\t=(i_1,\dots,i_d)$, let $\Lambda_\t$ denote the Young diagram
$(i_d-d,\dots,i_1-1)$. Then, using Lemma \ref{8.2}, we see that  the element $\t\in I_{d,n}$
is join-irreducible if and only if the associated Young diagram
$\Lambda_\t$ is a rectangle, i.e. all rows have the same number of boxes.
We also observe that  $\t$ is join-irreducible if and only if $X(\t)$ is nonsingular
(this is a consequence of the fact that $X(\t)$ is nonsingular if and only if
$\Lambda_\t$ is a rectangle, cf. \cite{LW}).
\end{rem}

\begin{prop}\label{13.9}
The restriction map $R\to R_{\ui}$ is surjective, and the kernel is generated as an ideal by
$\{p_{\uj}\mid\ui\not\ge\uj\}$
\end{prop}


\subsection{Evaluation of Pl\" ucker coordinates on the opposite big cell in $G/P_d$}
 Consider the morphim $\f_d:G\to \Bbb{P}(\w^dV)$, where
$\f_d=f_d\circ \te_d$, $\te_d$ being the natural projection $G\to G/P_d$. Then $p_{\uj}
(\f_d(g))$ is simply the minor of $g$ consisting of the first $d$ columns and the rows with
indices $j_1,\dots, j_d$. Now, denote by $Z_d$ the unipotent subgroup of $G$ generated by
$\{U_\a\mid\a\in R^-\setminus R_{P_d}^-\}$. We have, as in \S \ref{13.4}
$$Z_d=\left\{
\begin{pmatrix}
I_{d\times d}&0_{d\times (n-d)}\\
A_{(n-d)\times d}&I_{(n-d)\times (n-d)}
\end{pmatrix}\in G\right\}$$
As in \S \ref{13.4}, we identify $Z_d$ with the opposite big cell in $G/P_d$. Then, given $z\in
Z_d$,  the Pl\" ucker coordinate $p_{\uj}$ evaluated at $z$ is simply a certain minor of $A$,
which may be explicitly described as follows. Let $\uj=(j_1,\dots,j_d)$, and let $j_r$ be the
largest entry $\le d$. Let $\{k_1,\dots,k_{d-r}\}$ be the complement of $\{j_1,\dots,j_r\}$ in
$\{1,\dots,d\}$. Then this minor of $A$ is given by column indices $k_1,\dots k_{d-r}$, and row
indices $j_{r+1},\dots,j_d$ (here the rows of $A$ are indexed as $d+1,\dots,n$). Conversely,
given a minor of $A$, say,  with column indices $b_1,\dots,b_s$, and row indices
$i_{d-s+1},\dots,i_d$, it is the evaluation of the Pl\" ucker coordinate $p_{\ui}$ at $z$, where
$\ui=(i_1,\dots,i_d)$ may be described as follows: $\{i_1,\dots,i_{d-s}\}$ is the complement of
$\{b_1,\dots,b_s\}$ in
$\{1,\dots,d\}$, and $i_{d-s+1},\dots,i_d$ are simply the row indices (again, the rows of $A$ are
indexed as
$d+1,\dots,n$).

\subsection{Evaluation of the Pl\" ucker coordinates on the opposite big cell in
$G/Q$}\label{13.7} Consider
$$f:G\to G/Q\hookrightarrow G/P_{a_1}\times\dots\times
G/P_{a_k}\hookrightarrow\bold{P}_1\times\dots\times\bold{P}_k,$$ where
$\bold{P}_t=\Bbb{P}(\w^{a_t}V)$. Denoting the restriction of $f$ to $O^-$ also by just $f$, we
obtain an embedding $f:O^-\hookrightarrow \bold{P}_1\times\dots\times\bold{P}_k$, $O^-$ having
been identified with the opposite big cell in $G/Q$. For $z\in O^-$, the multi-Pl\" ucker
coordinates of $f(z)$ are simply all the $a_t\times a_t$ minors of $z$ with column indices
$\{1,\dots,a_t\}$, $1\le t\le k$.


\subsection{Equations defining multicones over Schubert varieties in $G/Q$}\label{13.10}
Let $X_Q(w)\subset G/Q$. Denoting $R$, $R_w$ as in \S \ref{12.6}, the kernel of the restriction
map
$R\to R_w$ is generated by the kernel of $R_1\to (R(w))_1$; but now, in view of Proposition
\ref{13.9}, this kernel is the span of
$$\{p_{\ui}\mid\ui\in I_{d,n},d=a_1,\dots,a_k,w^{(d)}\not\ge\ui\},$$
where $w^{(d)}$ is the $d$-tuple corresponding to the Schubert variety which is the image of
$X_Q(w)$ under the projection $G/Q\to G/P_{a_t}$, $1\le t\le k$.

\subsection{Ideal of the opposite cell in $X(w)$}
Let us denote $B^-e_{\text{id},Q}\cap X_Q(w)$ by just $A_w$. Then as in \S \ref{12.5}, we identify
$B^-e_{\text{id},Q}$ with the unipotent subgroup $O^-$ generated by $\{U_\a\mid\a\in R^-\setminus
R^-_Q\}$, and consider $A_w$ as a closed subvariety of $O^-$. In view of \S \ref{13.10}, we obtain
\begin{prop}
The ideal defining $A_w$ in $O^-$ is generated by
$$\{p_{\ui}\mid\ui\in I_{d,n},d=a_1,\dots,a_k,w^{(d)}\not\ge\ui\}$$
(here, for $z\in O^-$, $p_{\ui}(z)$ is as in \S \ref{13.7}).
\end{prop}


%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%% ladders, d=2

\section{Singular loci of  certain ladder determinantal varieties}\label{s14}
In this section we determine the singular loci of certain ladder determinantal varieties. Viewing
$X_{2,n}$ as a ladder determinantal variety, we  prove the conjecture of Section \ref{s11}, for
the case $d=2$.

We assume $d=2$, and view $\L=\{(i,j)\mid 1\le i<j\le n\}$ as being contained in $Y=\{(i,j)\mid
1\le i,j\le n\}$.
A Schubert variety $X(\t)$  in $G_{2,n}$ degenerates to $X(\L_\t)$, where
$\L_\t=\{\te\in\L\mid\te\le\t\}\subset \L$ (cf. \cite{GLdef}). The equations defining the variety
$X(\L_\t)$ are precisely the $ 2$ minors of $Y$ which are contained in $\L_\t$. Now we
look at a more general type of varieties.
\subsection{}
Let $Y=\{(b,a)\mid 1\le b\le n,\,1\le a\le n\}$.
Given $1\le b_1<\dots<b_l< n$, $1< a_1<\dots<a_1\le n$, we consider the subset $L$ of $Y$,
defined by
$$L=\{(b,a)\mid\text{ there exists } 1\le i\le l\text{ such that }b_i\le b\le n,\,1\le a\le
a_i\}.$$ We call $L$ an {\em one-sided ladder} in $Y$, defined by the {\em outer corners}
$\a_i=(b_i,a_i)$, $1\le i\le l$ (see Figure $1$). For $1\le i\le l$, let $L_i$ be the subset of
$Y$ defined by
$$L_i=\{(b,a)\mid b_i\le b\le n, 1\le a\le a_i\}.$$
Clearly, $L=\cup_{i=1}^l L_i$.

We also view $Y$ as the generic $n\times n$ matrix $(x_{ba})_{1\le b,a\le n}$, and we say that
$x_{ba}\in L$ if $(b,a)\in L$. Let $k[L]$ denote the polynomial ring $k[x_{ba}\mid x_{ba}\in L]$,
and let $\Bbb{A}(L)=\Bbb{A}^{\# L}$ be the associated affine space. Let $I(L)$ be the ideal in
$k[L]$ generated by the $ 2$ minors of $Y$ which are contained in $L$, and $X(L)\subset
\Bbb{A}(L)$ the variety defined by the ideal $I(L)$. A $ 2$ minor in $Y$ formed with two
consecutive rows and two consecutive columns will be  called a {\em solid minor}. We have (cf.
\cite{KS}):
\begin{lem}\label{14.2}
The codimension of $X(L)$ in $\Bbb{A}(L)$ is equal to the number of solid $ 2$ minors in
$L$.
\end{lem}
\begin{thm}\label{14.3}
Let $L$ be an one-sided ladder in $Y$ defined by the outer corners $\a_i=(b_i,a_i)$, with
$a_i\ge 2$ and $b_i\le n-1$, $1\le i\le l$. For each
$1\le i\le l$, let
$V_i=\{P=(P_\a)_{\a\in L}\in X(L)\mid P_\a=0\text{ for all }\a\in L_i\}$.
Then  the irreducible components of $\text{Sing\,}X(L)$ are precisely $V_i$, $1\le i\le l$.
\end{thm}
\begin{pf}
First, we prove that $V_i\subset\text{Sing\,}X(L)$, for all $1\le i\le l$. Fix $1\le i\le l$,
and let $P=(P_\a)_{\a\in L}\in V_i$. Let $\J$ be the Jacobian matrix of $X(L)$. Then the rows of
$\J$ are indexed by $\{ M\mid M\text{ is a }  2\text{ minor of }Y\text{ contained in }L\
\}$, and the columns are indexed by
$\{x_\a\mid\a\in L\}$. Let
$\J_P$ be the Jacobian matrix evaluated at $P$. Then the $(M,x_\a)$-th entry in $\J_P$ is equal
to
$\pm P_{\a'}$, where
$\a'$ is the element in $M$ which is neither in  the row, nor in the column containing $\a$, if
$x_\a$ appears in $M$, and $0$ otherwise. Since $P\in V_i$, the row indexed by a minor involving
$x_{\a_i}=x_{(b_i,a_i)}$ is $0$. Also, the column indexed by $x_{\a_i}$ is $0$.

Let $\J'$ be the matrix obtained from $\J_P$ by deleting the column indexed by $x_{\a_i}$, and the
rows indexed by minors involving $x_{\a_i}$. Then
\begin{equation*}
\text{rank\,}\J_P=\text{rank\,}\J',\tag{$1$}
\end{equation*}
since $\J'$ is obtained from $\J_P$ by deleting zero rows and columns.

Now consider the one-sided ladder $L'$ obtained from $L$ by deleting the element $\a_i$. Let
$P'=(P_\a)_{\a\in L'}$. Then $P'\in X(L')$, and $\J'$ is the Jacabian matrix of $X(L')$
evaluated at $P'$. Thus
\begin{equation*}
\text{rank\,}\J'\le\text{codim\,}_{\Bbb{A}(L')}X(L').\tag{$2$}
\end{equation*}
By hypothesis, $a_i\ge 2$ and $b_i\le n-1$, and hence there exist at least one solid minor in
$L$ not contained in $L'$, while every solid minor in $L'$ is also contained in $L$. Thus, using
Lemma \ref{14.2}, we deduce that
\begin{equation*}
\text{codim\,}_{\Bbb{A}(L')}X(L')<\text{codim\,}_{\Bbb{A}(L)}X(L).\tag{$3$}
\end{equation*}
Using $(1)$, $(2)$ and $(3)$, we deduce that
$$\text{rank\,}\J_P<\text{codim\,}_{\Bbb{A}(L)}X(L),$$
i.e. $P\in\text{Sing\,}X(L)$.

Next we prove that $\text{Sing\,}X(L)\subset \cup_{i=1}^lV_i$. Let $P\in X(L)\ \setminus\
\cup_{i=1}^lV_i$. For each $1\le i\le l$, we fix an element $\b_i\in L_i$ such that $P_{\b_i}\ne
0$. Let $\C$ be the set obtained from $L$ by deleting the elements appearing either in
the first column, or in the last row of $L$. Then $\#\C$ is equal to the number of solid
$ 2$ minors in
$L$, and by Lemma \ref{14.2}, we have
$$\#\C=\text{codim\,}_{\Bbb{A}(L)}X(L).$$
We have $\C=\dot{\cup}_{i=1}^l\C_i$, where $\C_i=\C\cap (L_i\setminus L_{i+1})$, for $1\le i\le
l$,
$L_{l+1}=\emptyset$. Let
$\T_l$ be the set of elements in $L_l$ not in the row or the column of
$\b_l$. Clearly, $\#\T_l=\#\C_l$. By (decreasing) induction on $i$, suppose that, for
some $i$, $1<i\le l$, the  sets $\T_i,\dots,\T_l$
 have been constructed, such that

$(1)_i$ $\T_j\subset L_j$, $i\le j\le l$,

$(2)_i$ the sets $\T_i,\dots,\T_l$ are pairwise disjoint,

$(3)_i$ $\#\T_j=\#\C_j$, $i\le j\le l$,

$(4)_i$ $\T_j$ contains no elements appearing in the column or in the row of $\b_j$, $i\le j\le
l$,

$(5)_i$ there exists a row in $L_i$ not containing any element from $\T_i\cup\dots\cup\T_l$.

We define the set $\T_{i-1}$ as follows. If
$\b_{i-1}\not\in L_{i-1}\setminus L_i$, then $\T_{i-1}$ is obtained from $L_{i-1}\setminus L_i$ by
deleting the elements in the column of $\b_{i-1}$. If $\b_{i-1}\in L_{i-1}\setminus L_i$, then
$\T_{i-1}$ is obtained as follows. Choose a row $R_i$ in $L_i$, as given by $(5)_i$. We set
$\T_{i-1}=(L_{i-1}\setminus L_i)\cup R_i\setminus A_{\b_{i-1}}$, where $A_{\b_{i-1}}$ is the set
of elements of $L$ in the row and column of $\b_{i-1}$.
 Clearly,
$\T_{i-1}\subset L_{i-1}$, the sets $\T_{i-1},\T_i,\dots,\T_l$ are pairwise disjoint,
$\#\T_{i-1}=\#\C_{i-1}$, and
$\T_{i-1}$ does not contain any element in the row or the column of $\b_{i-1}$. Also, there
exists a
 row in $L_{i-1}$ which does not contain any  element from
$\T_{i-1}\cup\T_i\cup\dots\cup\T_l$. Hence the sets $\T_{i-1},\T_i,\dots,\T_l$ satisfy
$(1)_{i-1}-(5)_{i-1}$. Thus  we obtain  pairwise disjoint sets
$\T_j\subset L_j$, $1\le j\le l$, such that
$\#\T_j=\#\C_j$, and $\T_j$ does not contain any element in the row or column of $\b_j$. Let
$\T=\cup_{i=1}^l\T_i$. Then $\#\T=\#\C$.

For $\t\in\T_i$, $1\le i\le l$, let $M^\t$ be the $ 2$ minor determined by $\t$ and
$\b_i$. Clearly, $M^\t\ne M^{\t'}$ for $\t,\,\t'\in\T$, $\t\ne\t'$.
Let $\J'$ be the submatrix of $\J_P$ given by rows indexed by $M^\t$'s and the columns indexed by
$x_\t$'s, with $\t\in\T$. We index the rows and columns of $\J'$  by the
elements in $\T$, and we arrange them increasingly, with respect to the lexicographic  order in
$Y$ (namely, $(b,a)>(b',a')$ if $b>b'$, or $b=b'$, $a>a'$).

Let us fix $\t\in\T$, say $\t\in\T_i$, $1\le i\le l$. Since
$\t$ is the only entry in $M^\t$ which belongs to $\T$, we deduce that in the $\t$-th row of $\J'$
all the entries are zero, except the one in the $\t$-th column, which is equal to $\pm P_{\b_i}$,
and hence it is nonzero.
Thus the matrix $\J'$ is diagonal, with nonzero diagonal entries. Therefore its rank is equal to
its size, which is $\#\T$. Since $\#\T$ is equal to $\#\C$, and hence  equal to the codimension of
$X(L)$, we deduce that
\begin{equation*}
\text{rank\,}\J'=\text{codim\,}_{\Bbb{A}(L)}X(L).\tag{$4$}
\end{equation*}
Since $\text{rank\,}\J'\le\text{rank\,}\J_P\le \text{codim\,}_{\Bbb{A}(L)}X(L)$, $(4)$ implies
that
$$\text{rank\,}\J_P=\text{codim\,}_{\Bbb{A}(L)}X(L),$$
i.e. $P\not\in\text{Sing\,}X(L)$.
Therefore we conclude that
$$\text{Sing\,}X(L)=\cup_{i=1}^lV_i.$$
Let $\L_i=L\setminus L_i$, $1\le i\le l$. Then $\L_i$ is a distributive lattice, and $V_i$ is
identified with $X(\L_i)$. Using \S \ref{5.13}, we deduce that $V_i$ is irreducible, $1\le i\le
l$. The fact that $V_i\not\subset V_j$ for $i\ne j$ is obvious. This completes the proof of the
theorem.
\end{pf}
The following theorem shows the validity of the Conjecture stated in Section \ref{s11}, for the
case $d=2$.
\begin{thm}
Let $\L_{i1}\subset I_{2,n}$, $1\le i\le n-3$, be as in Section \ref{s11}.  Then the irreducible
components of
$\text{Sing\,}X_{2,n}$ are precisely $X(\L_{i1})$, $1\le i\le n-3$.
\end{thm}
\begin{pf}
First observe that $\f_1=(1,n)$ is the only join and meet-irreducible element of Type II in
$I_{2,n}$, and the join and meet-irreducible elements of Type I are $\t_i=(i+1,i+2)$, $0\le i\le
n-2$ (note that $\t_0=\widehat{0}$, $\t_{n-2}=\widehat{1}$). We have $X_{2,n}=X(L)\times
\Bbb{A}^2$, where
$L=I_{2,n}\setminus\{\widehat{0},\widehat{1}\}$. Using Theorem \ref{14.3} for the ladder $L$,
we obtain that the irreducible components of $X(L)$ are $V_i$ ,
$1\le i\le n-3$ ($V_i$ being as defined in the statement of the Theorem). Thus the irreducible
components of
$X_{2,n}$ are precisely
$V_i\times
\Bbb{A}^2$, $1\le i\le n-3$. It is easily  seen that $X(\L_{i1})=V_i\times \Bbb{A}^2$, $1\le
i\le n-3$, and the result follows from this.
\end{pf}
\begin{rem}
We have $$\text{codim\,}_{X(L)}V_i=b_{i+1}-b_i+a_i-a_{i-1}+1,\quad 1\le i\le l$$
(here, $a_0=0$, $b_{l+1}=n$). In particular, taking  $L=I_{2,n}\setminus
\{\widehat{0},\widehat{1}\}$, we deduce that $\text{Sing\,}X_{2,n}$ is of pure codimension three
in $X_{d,n}$.
\end{rem}



%%%%%%%%%%%%%%%%%%%%%%%%%%
%%% schubert

\section{Relationship between ladder determinantal varieties and Schubert varieties}\label{s15}
Let $Y=(x_{ba})$, $1\le b, a\le n$ be a matrix of variables, and
$L\subset Y$ an one-sided ladder defined by the outer corners $(b_1,a_1),\dots,(b_l,a_l)$, with
$1\le b_1<\dots<b_l< n$, $1< a_1<\dots<a_l\le n$. We also assume that $n$ is large enough
such that $L$ lies below the main diagonal of $Y$, i.e. $b_i>a_i$ for all $1\le i\le l$.

Let $G=SL(n)$, and consider the parabolic subgroup $Q=P_{a_1}\cap\dots\cap P_{a_l}$ in $G$. Let
$O^-$ be the opposite big cell in $G/Q$. Let $X(L)\subset \Bbb{A}(L)$ be the the variety defined
by the vanishing of the $ 2$ minors in $L$, as in Section \ref{s14}. Let  $H$ be the one-sided
ladder defined by the outer corners $(a_i+1,a_i)$, $1\le i\le l$, and let $Z$ be the variety in
$\Bbb{A}(H)\simeq O^-$ defined by the vanishing of the
$2$ minors in $L$. Note that $Z\simeq X(L)\times \Bbb{A}(H\setminus L)$.

 Let $Y^-=(y_{ba})$, $1\le b,a\le n$, where
$$y_{ba}=
\begin{cases}
x_{ba},&\text{if } (b,a)\in H\\
1,&\text{if }b=a\\
0,&\text{otherwise}.
\end{cases}
$$
Note that, given $\t\in W^{a_i}$, $1\le i\le l$, the function $p_\t\res_{O^-}$ represents the
 determinant of the submatrix $T$ of $Y^-$ whose row indices are $\{\t(1),\dots,\t(a_i)\}$, and
column indices are $\{1,\dots, a_i\}$.

We shall now define an element $w\in W^Q$ such that $Z$ gets identified with
the opposite cell  in $X(w)$ (see also \cite{M}). We define
$w\in W^Q$ by specifying $w^{(i)}\in W^{a_i}$, where $\pi_i(X(w))=X(w^{(i)})$ under the projection
$\pi_i:G/Q\to G/P_{a_i}$, $1\le i\le l$.

Define $w^{(a_i)}$, $1\le i\le l$, inductively, as the maximal elements in $W^{a_i}$
with the following properties:

$(1)$ $w^{(a_i)}(a_i-1)=b_i-1$,

$(2)$ if $i>1$, then $w^{(a_i-1)}\subset w^{(a_i)}$.

Note that $w\in W^Q$, and it  is maximal for the property
$$w^{(a_i)}(a_i-1)<b_i,$$
for all $1\le i\le l$.

\begin{thm}\label{15.1}
The variety $Z$ identifies with the opposite  cell
in $X(w)$, i.e. $Z=X(w)\cap O^-$ (scheme theoretically).
\end{thm}
\begin{pf}
Let $f$ be a generator of $I(Z)$, i.e. $f=\det M$, for some $2\times 2$
matrix $M$ contained in $L$.  Let $r_1<r_2$ (resp. $c_1<c_2$) be the row (resp. column)
indices of $M$. Let $1\le i\le l$ smallest such that $M$ is contained in  $L_i$. Thus $r_1,r_2\ge
b_i>a_i$, $c_2>a_{i-1}$ (here $a_{i-1}=0$ if $i=1$). Let
$\t=\{1,\dots,a_i\}\setminus\{c_1,c_2\}\cup\{r_1,r_2\}$. Then $\t\in W^{a_i}$, and
 $p_{\t}\res_{O^-}=\det T$, where $T$ is the $a_i\times a_i$ submatrix of $Y^-$ whose row
indices are
$\{\t(1),\dots,\t(a_i)\}$, and column indices are $\{1,\dots,a_i\}$. Using Laplace expansion
with respect to the last two rows of $T$, we obtain
\begin{equation*}
\det T=\sum\pm\det N_{c',c''}\det M_{c',c''},\tag{$\ast$}
\end{equation*}
the sum being taken over all subsets $\{c',c''\}\subset\{1,\dots,a_i\}$, $c'\ne c''$, where
$N_{c',c''}$ is the
$(a_i-2)\times (a_i-2)$
 submatrix of $Y^-$ whose row indices are
$\{1,\dots,a_i\}\setminus\{c_1,c_2\}$ and column indices are $\{1,\dots,a_i\}\setminus
\{c',c''\}$, and $M_{c',c''}$ is the $2\times 2$ submatrix of $Y^-$ whose row indices are
$\{r_1,r_2\}$ and column indices are $\{c',c''\}$.  Note that
$M_{c_1,c_2}=M$, and $N_{c_1,c_2}$ is a lower triangular matrix, with all diagonal entries equal
to $1$, and hence $\det M$ appears in $(\ast)$, and its coefficient is $\pm 1$. Also note that
$N_{c',c''}$ is obtained from $N_{c_1,c_2}$ by replacing the columns with indices $c', c''$ by
the columns with indices $c_1, c_2$.

By decreasing induction with respect to the index $c_1$ of the first column of $M$, we prove that
$f=\det M$ can be written in the form
 $f=\sum g_\f p_\f\res_{O^-}$, with $\f\in W^{a_i}$,
$\{a_i+1,\dots,n\}\cap\{\f(1),\dots,\f(a_i)\}=\{r_1,r_2\}$, and $g_\f\in k[H]$.

If $c_1>a_{i-1}$, then  for
$\{c',c''\}\ne\{c_1,c_2\}$ we have $\det N_{c',c''}=0$, since at least one of $c_1,c_2$ is
an index for a column in $N_{c',c''}$, and all entries of this column are $0$. Thus, in this case
$(\ast)$ reduces to $\det T=\pm \det M$, i.e. $\det M=\pm p_\t\res_{O^-}$, with $\t\in
W^{a_i}$ such that
$\{a_i+1,\dots,n\}\cap\{\t(1),\dots,\t(a_i)\}=\{r_1,r_2\}$.

Let $c_1\le a_{i-1}$ such that the above statement is true for $c_1+1$.  First we observe that if
$c_2\not\in\{c',c''\}$, then $\det N_{c',c''}=0$, since in this case
$c_2$ is an index for a column in $N_{c',c''}$, and all entries of this column are $0$.
Let then  $c_2\in\{c',c''\}$, and let $\{c',c''\}=\{c,c_2\}$. Then $N_{c,c_2}$ is
obtained from $N_{c_1,c_2}$ by replacing the column with index  $c$ by the
column with index  $c_1$.  If $c<c_1$, then $N_{c,c_2}$ is still lower triangular, but the
diagonal entry in the column with index  $c_1$ is $0$, and hence $\det N_{c,c_2}=0$.
Therefore we obtain
\begin{equation*}
\det T=\pm\det M+\sum_{c\in\{c_1+1,\dots,a_i\}\setminus\{c_2\}}\pm\det N_{c,c_2}\det M_{c,c_2},
\end{equation*}
and hence
\begin{equation*}
f=\det M=\pm p_{\t}\res_{O^-}+\sum_{c\in\{c_1+1,\dots,a_i\}\setminus\{c_2\}}\pm\det N_{c,c_2}\det
M_{c,c_2}.
\end{equation*}

Using induction hypothesis for $M_{c,c_2}$,  we obtain $f=\sum g_\f p_\f\res_{O^-}$, with
$\f\in W^{a_i}$ such that $\{a_i+1,\dots,n\}\cap\{\f(1),\dots,\f(a_i)\}=\{r_1,r_2\}$, and
$g_\f\in k[H]$. In particular $\f(a_i-1)=r_1$. Since $r_1\ge b_i$, we deduce that
$\f(a_i-1)\ge b_i$. We have $w^{(a_i)}(a_i-1)=b_i-1$, and hence  $\f(a_i-1)>w^{(a_i)}(a_i-1)$.
This shows that $\f\not\le w^{(a_i)}$, and hence $p_\f\res_{O^-}\in I(X(w)\cap O^-)$. Therefore
$f\in I(X(w)\cap O^-)$.

Let now $g$ be a generator of the ideal $I(X(w)\cap O^-)$, i.e. $g=p_\t\res_{O^-}$, with
$\t\in W^{a_i}$ for some $1\le i\le l$, such that $\t\not\le w^{a_i}$ (cf. \S \ref{13.5}). Since
$w^{(a_i)}$ consists of several blocks of consecutive integers ending with $b_t-1$ at the
$(a_t-1)$-th place, for some $t$'s in $\{1,\dots,i\}$, and a last block ending with $n$ at the
$a_i$-th place, it follows that there exists a $t\in\{1,\dots,i\}$ such that $\t(a_t-1)\ge b_t$.
As above, the function $p_\t\res_{O^-}$ represents the determinant of the submatrix $T$ of $Y^-$
whose row indices are
$\{\t(1),\dots,\t(a_i)\}$, and column indices are $\{1,\dots, a_i\}$. Using Laplace expansion with
respect to the first $a_t$ columns,we have $\det T=\sum_p\det A_p\det B_p$, where $A_p$
(resp.
$B_p$) is an $a_t\times a_t$ (resp. $(a_i-a_t)\times (a_i-a_t)$) matrix.  Clearly, all the column
indices of $A_p$ are $\le a_t$, and since
$\t(a_t-1)\ge b_t$, at least $2$ row indices of $A_p$ are $\ge b_t$. Using Laplace expansion
for $A_p$ with respect to $2$ rows whose indices are $\ge b_t$, we obtain $\det A_p=\sum_q\det
C_q\det D_q$, where $C_q$ (resp. $D_q$) is a $2$ (resp. $a_t-2$) minor, with $C_q$ contained in
$L_t\subset L$. This shows that $p_{\t}\res_{O^-}\in I(Z)$.
 This completes the proof.
\end{pf}

\begin{cor}
The variety $X(L)$ is normal, Cohen-Macaulay, and has rational singularities.
\end{cor}
This follows from Theorem \ref{15.1}, and the fact that Schubert varieties are normal,
Cohen-Maculay, and have rational singularities (cf. \cite{R1}, \cite{R2}, and \cite{KR}).

Let us fix $j\in\{1,\dots,l\}$. We shall now define $\te_j\in W^Q$ such that $Z_j=V_j\times
\Bbb{A}(H\setminus L)$ gets identified with the opposite cell in $X(\te_j)$.

 For $i<j$, let
$\te_j^{(a_i)}=w^{(a_i)}\setminus\{n\}\cup\{b_j-1\}$.

For $i=j$, let $\te_j^{(a_j)}=w^{(a_j)}\setminus\{n\}\cup\{x_j\}$, where $x_j$ is the maximal
element in
$\{1,\dots,b_j-1\}\setminus w^{(a_j)}$.

For $i>j$, let
$$\te_j^{(a_i)}=
\begin{cases}
w^{(a_i)},&\text{if }x_j\in w^{(a_i)}\\
w^{(a_i)}\setminus\{y_i\}\cup\{x_j\},&\text{if }x_j\not\in w^{(a_i)},
\end{cases}
$$
where $y_i$ is the minimal element in $w^{(a_i)}\setminus \te_j^{(a_{i-1})}$.

\begin{lem}\label{15.2}
We have $\te_j\le w$. Further, $\te_j$ is the maximal
element $\t\in W^Q$, $\t\le w^{\text{max}}$, such that $n\not\in\t^{(a_j)}$, and
$n\in\t^{(a_{j+1})}$.
\end{lem}
The assertion is clear from the definition of $\te_j$.

\begin{thm}\label{15.4}
The subvariety $Z_j\subset Z$ gets identified with the opposite  cell in $X(\te_j)$, i.e.
$Z_j=X(\te_j)\cap O^-$ (scheme theoretically).
\end{thm}
\begin{pf}
Let $f$ be a generator of $I(Z_j)$. If $f\in I(Z)$, then, in view of Theorem \ref{15.1} we have
$f\in I(X(w)\cap O^-)\subset I(X(\te_j)\cap O^-)$ (since $w\ge\te_j$), and there is nothing to
prove. Assume that
$f\not\in I(Z)$; then $f=x_\a$, for some $\a=(b,a)\in L_j$. Then $f$ can be written as
$p_\f\res_{O^-}$, with $\f\in W^{a_j}$, such that
$\{b\}=\{a_j+1,\dots,n\}\cap\{\f(1),\dots,\f(a_j)\}$, and
$\{a\}=\{1,\dots,a_j\}\setminus\{\f(1),\dots,\f(a_i)\}$. Thus  $\f(a_j)=b$, and since $\a\in L_j$,
we have $b\ge b_j$. Therefore $\f(a_j)\ge b_j$. But
$\te_j^{(a_j)}(a_j)=b_j-1$, and hence $\f(a_j)>\te_j^{(a_j)}(a_j)$. This shows that
$\f\not\le\te_j^{(a_j)}$, and therefore $f\in I(X(\te_j)\cap O^-)$.

Let now $g$ be a generator of the ideal $I(X(\te_j)\cap O^-)$, i.e. $g=p_\t\res_{O^-}$, with
$\t\in W^{(a_i)}$, for some $1\le i\le l$, such that $\t\not\le \te_j^{(a_i)}$.


First assume that $i\le j$. Then $\te_j^{(a_i)}$ consists of several blocks of consecutive
integers ending with $b_t-1$ at the $(a_t-1)$-th place, for some $t$'s in $\{1,\dots,i-1\}$,
and a last block ending with $b_j-1$ at the $a_i$-th place. The condition $\t\not\le\te_j^{(a_i)}$
implies that either there exists $t\in\{1,\dots,i-1\}$ such that $\t(a_t-1)\ge b_t$, or
$\t(a_i)\ge b_j$. In the first case we have $\t\not\le w^{(a_i)}$, and hence $p_\t\res_{O^-}\in
I(X(w)\cap O^-)=I(Z)\subset I(Z_j)$. Suppose now that $\t(a_i)\ge b_j$. The function
$p_\t\res_{O^-}$ represents the determinant of the submatrix $T$ of $Y^-$ whose row indices are
$\{\t(1),\dots,\t(a_i)\}$, and column indices are
$\{1,\dots,a_i\}$. Obviously, all column indices are $\le a_j$. On the other hand, since
$\t(a_i)\ge b_j$, the last row of $T$ is contained in $L_i$, and expanding $T$
along this row, we deduce $p_\t\res_{O^-}\in I(Z_j)$.

Assume now that $i>j$.
If $\te_j^{(a_i)}=w^{(a_i)}$, then $\t\not\le w^{(a_i)}$, and hence
$p_\t\res_{O^-}\in I(X(w)\cap O^-)=I(Z)\subset I(Z_j)$, and there is nothing to prove. Suppose
that
$\te_j^{(a_i)}\ne w^{(a_i)}$. Then $\te_j^{(a_i)}$ consists of several blocks of consecutive
integers ending with $b_t-1$ at the $(a_t-1)$-th place, for some $t$'s in
$\{1,\dots,i\}\setminus\{j\}$, a block ending with $b_j-1$ at the $a_j$-th place, and a last
block ending with $n$ at the $a_i$-th place. The
condition
$\t\not\le
\te_j^{(a_i)}$ implies that either     there exists $t\in\{1,\dots,i\}\setminus\{j\}$ such that
$\t(a_t-1)\ge b_t$, or
$\t(a_j)\ge b_j$. In the first case we have $\t\not\le w^{(a_i)}$, and hence $p_\t\res_{O^-}\in
I(X(w)\cap O^-)=I(Z)\subset I(Z_j)$. In the second case,
the function  $p_\t\res_{O^-}$ represents the determinant of the submatrix $T$ of $Y^-$ whose
row indices are
$\{\t(1),\dots,\t(a_i)\}$, and column indices are $\{1,\dots, a_i\}$. Using Laplace expansion with
respect to the first $a_j$ columns, we have $\det T=\sum_p\det A_p\det B_p$, where $A_p$
(resp.
$B_p$) is an $a_j\times a_j$ (resp. $(a_i-a_j)\times (a_i-a_j)$) matrix.  Clearly, all the column
indices of $A_p$ are $\le a_j$, and since
$\t(a_j)\ge b_j$, at least one row index of $A_p$ is $\ge b_j$. Using Laplace expansion
for $A_p$ with respect to a row with index  $\ge b_j$, we obtain $\det A_p=\sum_qC_q \det
D_q$, where $C_q$'s are entries of a row of $A_p$, contained in $L_j\subset L$. This shows that
$p_{\t}\res_{O^-}\in I(Z_j)$.
\end{pf}

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

\section{Validity of the conjecture in $[17]$}\label{s16}
Let $G=SL(n)$. In this section we prove  the conjecture in \cite{LS} on the irreducible components
of the singular locus for a class of Schubert varieties, namely the pull-backs
$\pi^{-1}(X_Q(w))$ under
$\pi :G/B\to G/Q$, where $w$ and $Q$ are as in Section \ref{s15}. We have $\pi
^{-1}(X_Q(w))=X_B(w^{\text{max}})$, where $w^{\text{max}}$, as a permutation, is given by
$w^{(a_1)}$ arranged in descending order, followed by $w^{(a_2)}\setminus w^{(a_1)}$, etc.. We
shall refer to the set
$w^{(a_i)}\setminus w^{(a_{i-1})}$, $1\le i\le l+1$, arranged in descending order, as the
$i$-th block in $w^{\text{max}}$ (here, $w^{(a_0)}=\emptyset$, and $w^{(a_{l+1})}$ is the set
$\{1,\dots,n\}\setminus w^{(a_l)}$ arranged in descending order).
\begin{rem} \label{16.1}
All of the entries in the $i$-th block in $w^{\text{max}}$ are $\le
b_i-1$,
$2\le i\le l$.
\end{rem}

Let $\eta\in W$. We shall denote $X_B(\eta)$ by just $X(\eta)$.
 We first recall
the criterion given in \cite{LS} for $X(\eta)$ to be singular.
\begin{thm}
Let $\eta=(a_1\dots a_n)\in\cal{S}_n$. Then $X(\eta)$ is singular if and only if there exist
$i,j,k,m$, $1\le i<j<k<m\le n$ such that
$$\text{either }a_k<a_m<a_i<a_j\,\text{ or }a_m<a_j<a_k<a_i\, .$$
\end{thm}

\subsection{The set $E_{\eta}$}\label{16.3}
We next recall the conjecture in \cite{LS} on the irreducible components of
$\text{Sing\,}X(\eta)$.

Let $\eta=(a_1\dots a_n)\in\cal{S}_n$. Let $E_{\eta}$ be the set of all $\t'\le\eta$ 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 $\t'=(b_1\dots b_n)$, 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. $\eta'$) is the element obtained from $\eta$ (resp. $\t'$) 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 $\t'\ge\t$ and $\eta'\le\eta$.

\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 $\t'=(b_1\dots b_n)$, 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. $\eta'$) is the element obtained from $\eta$ (resp. $\t'$) 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 $\t'\ge\t$ and $\eta'\le\eta$.
\begin{conj}
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 $E_{\eta}$.
\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 occurance 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
occurance in} $\eta$. Let $(d,b,c,a)$ (resp. $(c',d',a',b')$) be a bad occurance of Type  I
(resp. Type II), where $a<b<c<d$ (resp. $a'<b'<c'<d'$). Let $\te$, $\te'$ be both $\le w$.
Further, let $b,a,d,c$ (resp. $a',c',b',d'$) appear in that order in $\te$ (resp. $\te'$). By
abuse of language, we shall refer to $(b,a,d,c)$ (resp. $(a',c',b',d')$) as a bad occurance in
$\te$ (resp. $\te'$) corresponding to the bad occurance $(d,b,c,a)$ (resp. $(c',d',a',b')$) in
$\eta$.



\begin{lem}\label{16.5}
Let $w$ be as in Section \ref{s15}. Then any bad occurance in $X(w^{\text{max}})$ is
of Type I.
\end{lem}
\begin{pf}
Let $w^{\text{max}}=(a_1\dots a_n)$. Assume that $(c,d,a,b)$ is a bad occurance of Type II in
$w^{\text{max}}$, where $a<b<c<d$. Clearly, $c$ and $d$ (resp. $a$ and $b$) cannot both
appear in the same block , in view of the description of $w^{\text{max}}$. Let then
$c,d,a,b$ appear in the $h$-th, $i$-th, $j$-th, $k$-th blocks respectively, where $h<i\le j<k$.
This implies that $a<b<c<d\le b_i-1$ (cf. Remark \ref{16.1}). But now, $a$ and $b$ are both
$<b_i-1$, and they both appear after $b_i-1$; further, $a$ appears before $b$ in
$w^{\text{max}}$, which is not possible by the construction of $w^{\text{max}}$ (note that
$a<b$). The required result follows from this.
\end{pf}
\begin{rem}\label{16.6}
Of course, there are several bad occurances in $w^{\text{max}}$ of Type I. For example, fix some
$j$, $1\le j\le l$. Then  $b_{j+1}-1$ belongs to the $(j+1)$-th block of $w^{\text{max}}$, and
$b_{j+1}-1\ge b_j$ (here, if $j=l$, then $b_{j+1}-1=n-1$). Let $z_{j+1}$ be the smallest entry
$\ge b_j$, in the $(j+1)$-th block of $w^{\text{max}}$. Observe that
$$z_{j+1}=
\begin{cases}
y_{j+1},&\text{if }x_j\not\in w^{(a_{i+1})}\\
b_j,&\text{if }x_j\in w^{(a_{i+1})}
\end{cases}
$$
(here, $x_j$, $y_{j+1}$ are as in Section \ref{s15}). Note that if $x_j\in w^{(a_{i+1})}$, then
$b_j$ also belongs to $w^{(a_{i+1})}$). Take $d=n$, $b=b_j-1$, $c=z_{i+1}$, $a=x_j$. Then
$d,b,c,a$ occur in the $1$-st, $j$-th, $(j+1)$-th, $m$-th blocks respectively, where $m\ge j+1$.
This provides an example of a Type I bad occurance in $w^{\text{max}}$.
\end{rem}
\begin{lem}\label{16.7}
Let $d,b,c,a$ be a bad occurance in $w^{\text{max}}$, where $a<b<c<d$. Assume that $b$ belongs
to the $i$-th block, for some $i$ (note that $i\le l$, since $b<c$). Then

$(1)$ $b\le b_i-1$

$(2)$ $d=n$.
\end{lem}
\begin{pf}
Let $d,b,c,a$ occur in the $h$-th, $i$-th, $j$-th, $m$-th blocks respectively in
$w^{\text{max}}$, where $h\le i< j\le m$. First observe that $b<n$, since $b<c<d\le n$. Hence,
if $i=1$, then $b\le b_1-1$. If $i\ge 2$, again $b\le b_i-1$ (cf. Remark \ref{16.1}).
\begin{claim}
$d>b_i-1$.
\end{claim}
\begin{pf}
Assume that $d\le b_i-1$. Then assumption implies $c<b_i-1$ (since $c<d$). Now both $c$ and
$b$ are $<b_i-1$, and $b$ belongs to the $i$-th block  in $w^{\text{max}}$. This implies that
$c$ should occur before $b$, which is not possible.
Hence our assumption is wrong, and the claim follows.
\end{pf}
Note that the Claim in fact implies that $d=n$ (and $h=1$).
\end{pf}
Let $\te\in SL(n)$ such that $\text{Sing\,}X(\te)\ne\emptyset$.
\begin{prop}\label{16.8}
The maximal elements in $E_{w^{\text{max}}}$ are precisely $\te_j^{\text{max}}$, $1\le i\le l$
(here $E_{w^{\text{max}}}$ is as in \S \ref{16.3}).
\end{prop}
\begin{pf}
We first observe that $\te_j^{\text{max}}\in E_{w^{\text{max}}}$; for, corresponding to the
bad occurance $d=n$, $b=b_j-1$, $c=z_{j+1}$, $a=x_j$ (cf. Remark \ref{16.6}), we have the bad
occurance
$(b,a,d,c)$ (note that $b,a,d,c$ occur in that order in $\te_j^{\text{max}}$. Let us denote
$\te_j^{\text{max}}$ by $\t'$. Let $w'$ (resp. $\t$) be the element of $\cal{S}_n$ obtained from
$\t'$ (resp. $w$) by replacing $b,a,d,c$ (resp. $d,b,c,a$) respectively by $d,b,c,a$ (resp.
$b,a,d,c$). Then clearly $\t\le\t'$, and $w'\le w$.

Let now $\t'\in E_{w^{\text{max}}}$. Then we have a bad occurance in $\t'$, which has to be of the
form $(b,a,d,c)$, $a<b<c<d$,  corresponding to the occurance $(d,b,c,a)$ in $w^{\text{max}}$ (cf.
Lemma \ref{16.5}). We have $d=n$ (cf. Lemma \ref{16.7}).
\begin{claim}
$d\not\in\t'{}^{(a_1)}$
\end{claim}
\begin{pf}
Assume that $d\in\t'{}^{(a_1)}$. This implies  that $b,a\in\t'{}^{(a_1)}$. Let $w'$ be the element
obtained from $\t'$ by replacing $b,a,d,c$ respectively by $d,b,c,a$. Now $w'\le w$ (cf. \S
\ref{16.3}) implies, in particular, that $w'{}^{(a_1)}\le w^{(a_1)}$. Hence we obtain that $b$ and
$c$ are both $\le b_1-1$. Now in $w^{\text{max}}$, we have, $c\not\in w^{(a_1)}$, and $c$ appears
after
$b$. But this is not possible (by the construction of $w^{\text{max}}$). The claim now follows.
\end{pf}
Now the claim implies that there exists an $r>1$ such that $d\in\t^{(a_r)}$,
$d\not\in\t^{(a_{r-1})}$. Hence    $\t'\le\te_{r-1}^{\text{max}}$ (cf. Lemma \ref{15.2}). The
required result now follows from this
\end{pf}
\begin{thm}\label{16.9}
The conjecture \ref{16.3} holds for $X(w^{\text{max}})$.
\end{thm}
\begin{pf}
In view of Theorems \ref{14.3}, \ref{15.1}, and \ref{15.4} $\text{Sing\,}X(\te_j^{\text{max}})$,
$1\le j\le l$ are precisely  the irreducible components of $X(w^{\text{max}})$. On the other
hand, we have (cf. Proposition \ref{16.8}) that the maximal elements in $E_{w^{\text{max}}}$ are
precisely
$\te_j^{\text{max}}$, $1\le j\le l$. Hence the irreducible components of
$\text{Sing\,}X(w^{\text{max}})$ are precisely
$\{X(\te)\mid\te\in E_{w^{\text{max}}}\}$. Thus the conjecture holds for $X(w^{\text{max}})$.
\end{pf}



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\end{document}