diagonal coordinates are $1$, and $W=\mathcal{S}_{n}$ is the Weyl group of
$G$.

###  5.1. Grassmannians, the flag variety and Schubert cells

We recall a few well-know facts about the flag variety and its decomposition
in Schubert cells from the computational point of view. We also set the
notation for the rest of the section.

######  Definition 5.1.1.

For $1\leq d\leq n$, the _Grassmannian_ over $k$ of $d$-planes in the
$n$-space is the set of linear subspaces of dimension $d$ in $k^{n}$. It is
noted $\operatorname{Gr}_{d,n/k}$.

For $V\in\operatorname{Gr}_{d,n/k}$, let $(v_{1},\ldots,v_{d})$ be a basis of
$V$. Then $v_{1}\wedge\ldots\wedge v_{d}$ is a non-zero element of
$\bigwedge^{d}k^{n}$. Choosing a different basis would change this alternate
product by a multiplicative factor (the determinant of one basis in the
other), therefore the line spanned by $v_{1}\wedge\ldots\wedge v_{d}$ in
$\bigwedge^{d}k^{n}$ is a well-defined element of
$\mathbb{P}(\bigwedge^{d}k^{n})$. One can also recover $V$ from this line by
the property that a vector $v$ is in $V$ if and only if $v\wedge
v_{1}\wedge\ldots\wedge v_{d}=0$ in $\bigwedge^{d+1}k^{n}$.

######  Definition 5.1.2.

The _Plücker embedding_ is the map

$\displaystyle\operatorname{Gr}_{d,n/k}$
$\displaystyle\longrightarrow\;\mathbb{P}\Big{(}\bigwedge^{d}k^{n}\Big{)}$
$\displaystyle\left\langle v_{1},\ldots,v_{d}\right\rangle$
$\displaystyle\longmapsto\;k\cdot(v_{1}\wedge\ldots\wedge v_{d})$

where $(v_{1},\ldots,v_{d})$ is any linearly independent family in $k^{n}$.

######  Remark 5.1.3.

Proposition 5.1.5 below together with the discussion above Definition 5.1.2
show that the Grassmannian is a projective variety and the Plücker embedding a
closed immersion.

We write any point $x\in\mathbb{P}(\bigwedge^{d}k^{n})$ as

$x=[x_{i_{1}\ldots i_{d}}]_{1\leq i_{1}<\ldots<i_{d}\leq n}$

using the projective coordinates along the standard basis
$(e_{i_{1}}\wedge\ldots\wedge e_{i_{d}})_{1\leq i_{1}<\ldots<i_{d}\leq n}$ of
$\bigwedge^{d}k^{n}$, where $(e_{i})_{1\leq i\leq n}$ is the standard basis of
$k^{n}$. It is sometimes convenient to use the variables $x_{i_{1}\ldots
i_{d}}$ for any sequence $(i_{1},\ldots,i_{d})$ of integers between $1$ and
$n$; they satisfy the relations $x_{i_{\sigma(1)}\ldots
i_{\sigma(d)}}=\operatorname{sgn}(\sigma)x_{i_{1}\ldots i_{d}}$ for any
$\sigma\in\mathcal{S}_{d}$ and $x_{i_{1},\ldots,i_{d}}=0$ whenever there are
$1\leq k<l\leq d$ such that $i_{k}=i_{l}$.

######  Definition 5.1.4.

The _Plücker coordinates_ of a point in $\operatorname{Gr}_{d,n/k}$ are the
projective coordinates $[x_{i_{1}\ldots i_{d}}]_{1\leq i_{1}<\ldots<i_{d}\leq
n}$ of its image under the Plücker embedding.

Let $(v_{1},\ldots,v_{d})$ be a linearly independent family in $k^{n}$, and
for any $1\leq j\leq d$ write the coordinates of $v_{j}$ along the standard
basis as $(v_{ij})_{1\leq i\leq n}$. Then the Plücker coordinate
$x_{i_{1}\ldots i_{d}}$ of the span $\langle v_{1},\ldots,v_{d}\rangle$ can be
computed as the rank $d$ minor of the matrix $(v_{ij})_{1\leq i\leq n,1\leq
j\leq n}$ along the rows $i_{1}$,…,$i_{d}$ and columns $1,\ldots,d$.

######  Proposition 5.1.5.

The Plücker coordinates satisfy the following quadratic relations, called the
_Plücker relations in dimension $d$_:

(5.1) $\sum_{k=1}^{d+1}(-1)^{k}x_{i_{1}\ldots
i_{d-1}j_{k}}x_{j_{1}\ldots\widehat{j_{k}}\ldots j_{d+1}}=0$

where $(i_{1},\ldots,i_{d-1})$ and $(j_{1},\ldots,j_{d+1})$ are any increasing
sequences of indices and the notation $\widehat{j}$ indicates that the index
$j$ is omitted. These relations define precisely the points in
$\mathbb{P}(\bigwedge^{d}k^{n})$ that arise from the Grassmannian.

###### Proof.

See for example [17, Thm. 1] or [10, §9.1 Lemma 1]. ∎

In the same vein as Proposition 5.1.5, one can express the property that a
subspace of $k^{n}$ is included into another.

######  Proposition 5.1.6.

Let $1\leq d<d^{\prime}\leq n$ be two integers. The set of pairs
$(V,V^{\prime})\in\operatorname{Gr}_{d,n/k}\times\operatorname{Gr}_{d^{\prime},n/k}$
such that $V\subset V^{\prime}$ is identified with the set of points

$\left([x_{i_{1},\ldots,i_{d}}]_{1\leq i_{1}<\ldots<i_{d}\leq
n},[x^{\prime}_{i_{1},\ldots,i_{d^{\prime}}}]_{1\leq
i_{1}<\ldots<i_{d^{\prime}}\leq
n}\right)\in\mathbb{P}\Big{(}\bigwedge^{d}k^{n}\Big{)}\times\mathbb{P}\Big{(}\bigwedge^{d^{\prime}}k^{n}\Big{)}$

which satisfy both sets of Plücker relations and the following additional
quadratic relations:

(5.2) $\sum_{k=1}^{d^{\prime}+1}(-1)^{k}x_{i_{1}\ldots
i_{d-1}j_{k}}x^{\prime}_{j_{1}\ldots\widehat{j_{k}}\ldots j_{d^{\prime}+1}}=0$

where $(i_{1},\ldots,i_{d-1})$ and $(j_{1},\ldots,j_{d^{\prime}+1})$ are any
increasing sequences of indices. These relations are called the _incidence
relations in dimensions $(d,d^{\prime})$_.

###### Proof.

See [10, §9.1 Lemma 2]. ∎

######  Definition 5.1.7.

A _flag_ in $k^{n}$ is a strictly increasing sequence $(V_{1},\ldots,V_{n-1})$
of proper subspaces of $k^{n}$. The _flag variety over $k$_ is the set of
flags and it is noted $\operatorname{Flag}_{n/k}$.

There is an obvious map
$\operatorname{Flag}_{n/k}\lhook\joinrel\relbar\joinrel\rightarrow\prod_{d=1}^{n-1}\operatorname{Gr}_{d,n/k}$
which, when composed with the product of the Plücker embeddings, gives the
embedding

(5.3)
$\operatorname{Flag}_{n/k}\lhook\joinrel\relbar\joinrel\rightarrow\prod_{d=1}^{n-1}\mathbb{P}\Big{(}\bigwedge^{d}k^{n}\Big{)}\,.$

This realises the flag variety as the projective subvariety of
$\prod_{d=1}^{n-1}\mathbb{P}(\bigwedge^{d}k^{n})$ defined by the Plücker
relations and the incidence relations (note that the Segre embedding [25,
Lemma 01WD] indeed makes $\prod_{d=1}^{n-1}\mathbb{P}(\bigwedge^{d}k^{n})$
projective).

We identify $\operatorname{Flag}_{n/k}$ with the algebraic group $G/B$ in the
standard way. As in the first paragraph of §3.1, we have the decomposition
$G/B=\coprod_{w\in\mathcal{S}_{n}}BwB/B$. The following is well-known (see for
example [10, §10.5 Exercise 11]):

######  Proposition 5.1.8.

The cell $BwB/B$ is the set of points in $G/B$ that satisfy the following
equations on Plücker coordinates, for all $d\in\\{1,\ldots,n-1\\}$:

(5.4) $x_{w(1)\ldots w(d)}\neq 0\quad\text{and}\quad x_{i_{1}\ldots
i_{d}}=0\quad\text{if}\quad\\{i_{1},\ldots,i_{d}\\}\not\preceq\\{w(1),\ldots,w(d)\\}$

where $\preceq$ is the order defined in Definition 2.4.2.

###  5.2. Explicit equations for $\widetilde{X}$ and $\widetilde{V}_{w}$

We give a description of the algebraic varieties $\widetilde{X}$ and
$\widetilde{V}_{w}$ of §3.2 as locally closed subvarieties of a projective
space defined by explicit equations.

Consider the morphism

$\displaystyle G\times\mathfrak{b}$
$\displaystyle\longrightarrow\;G/B\times\mathfrak{g}$ $\displaystyle(g,\psi)$
$\displaystyle\longmapsto\;(gB,\operatorname{Ad}(g)\psi).$

It induces an injection
$G\times^{B}\mathfrak{b}\lhook\joinrel\relbar\joinrel\rightarrow
G/B\times\mathfrak{g}$, which restricts to the isomorphism

(5.5)
$\widetilde{X}\stackrel{{\scriptstyle\sim}}{{\smash{\longrightarrow}\rule{0.0pt}{1.72218pt}}}\left\\{(gB,\psi)\in
G/B\times\mathfrak{b}\mathrel{}\middle|\mathrel{}\operatorname{Ad}(g^{-1})\psi\in\mathfrak{b}\right\\}$

between $\widetilde{X}$ and a closed subvariety of $G/B\times\mathfrak{b}$.
From now on, we use this isomorphism (5.5) as the definition of
$\widetilde{X}$. This way, the maps
$\widetilde{\pi}\colon\widetilde{X}\longrightarrow G/B$ and
$q\colon\widetilde{X}\longrightarrow\mathfrak{b}$ also introduced in §3.2, are
the projections on the first and second coordinate respectively. For
$w\in\mathcal{S}_{n}$ we have the cell

$\widetilde{V}_{w}=\left\\{(gB,\psi)\in\widetilde{X}\mathrel{}\middle|\mathrel{}g\in
BwB/B\right\\}$

and its Zariski-closure $\widetilde{X}_{w}$, which give the cell decomposition
$\widetilde{X}=\coprod_{w\in\mathcal{S}_{n}}\widetilde{V}_{w}$.

A point $(gB,\psi)\in\widetilde{X}$ will be identified by the projective
coordinates $x_{i_{1}\ldots i_{d}}$ of the flag $gB$ seen inside
$\prod_{d=1}^{n-1}\mathbb{P}(\bigwedge^{d}k^{n})$ by (5.3), and by the affine
coordinates $(u_{ij})_{i\leq j}$ given by the matrix coefficients of $\psi$.
We sometimes write $t_{i}$ instead of $u_{ii}$. We view $\psi$ as the matrix
of an endomorphism $u$ of $k^{n}$ written relatively to the standard basis
($u$ is upper triangular). We view $g$ as a flag $(V_{1},\ldots,V_{n-1})$, so
that $V_{d}$ is the span of the first $d$ columns of $g$.

######  Lemma 5.2.1.

Suppose that $u$ is an isomorphism. Then
$\operatorname{Ad}(g^{-1})\psi\in\mathfrak{b}$ if and only if, for each $1\leq
d\leq n-1$, the point $x_{d}=\left(x_{i_{1}\ldots
i_{d}}\right)_{i_{1}<\ldots<i_{d}}\in k^{\binom{n}{d}}$ is colinear to

$u(x_{d})\coloneqq\left(\sum_{j_{1}<\ldots<j_{d}}\Delta^{j_{1}\ldots
j_{d}}_{i_{1}\ldots i_{d}}(u)x_{j_{1}\ldots
j_{d}}\right)_{i_{1}<\ldots<i_{d}}\in k^{\binom{n}{d}}$

where $\Delta^{j_{1}\ldots j_{d}}_{i_{1}\ldots i_{d}}(u)$ is the minor of $u$
associated to rows $i_{1},\ldots,i_{d}$ and columns $j_{1},\ldots,j_{d}$. This
amounts to $\binom{n}{d}\left(\binom{n}{d}-1\right)/2$ equations.

###### Proof.

$\operatorname{Ad}(g^{-1})\psi$ is the matrix of $u$ relative to the basis
given by the columns of $g$. It is upper triangular if and only if each
$V_{d}$ is stable by $u$. Since $u$ is an isomorphism, $u(V_{d})$ is of
dimension $d$ and we can consider its Plücker coordinates: we specifically
want that $u(V_{d})$ and $V_{d}$ have the same image in $\bigwedge^{d}k^{n}$
under the Plücker embedding. The image of $V_{d}$ is the line spanned by the
vector

$x_{d}=\sum_{i_{1}<\ldots<i_{d}}x_{i_{1}\ldots
i_{d}}e_{i_{1}}\wedge\ldots\wedge e_{i_{d}}\quad\in\bigwedge^{d}k^{n}$

and a quick calculation shows that the image of $u(V_{d})$ is the line spanned
by

$u(x_{d})=\sum_{i_{1}<\ldots<i_{d}}\left(\sum_{j_{1}<\ldots<j_{d}}\Delta^{j_{1}\ldots
j_{d}}_{i_{1}\ldots i_{d}}(u)x_{j_{1}\ldots
j_{d}}\right)e_{i_{1}}\wedge\ldots\wedge
e_{i_{d}}\quad\in\bigwedge^{d}k^{n}\,.\qed$

######  Remark 5.2.2.

Note that, since $u$ is upper triangular, we have $\Delta^{j_{1}\ldots
j_{d}}_{i_{1}\ldots i_{d}}(u)=0$ unless
$\\{i_{1},\ldots,i_{d}\\}\preceq\\{j_{1},\ldots,j_{d}\\}$ for the partial
order of Definition 2.4.2. Also note that $\Delta^{i_{1}\ldots
i_{d}}_{i_{1}\ldots i_{d}}(u)=t_{i_{1}}\ldots t_{i_{d}}$.

######  Lemma 5.2.3.

Let $A$ be the polynomial ring in the coordinates of $(gB/B,\psi)$, that is
$A=\mathbb{Z}[x_{i_{1}\ldots i_{d}},u_{kl},t_{m}]_{1\leq d<n,\>1\leq
i_{1}<\ldots<i_{d}\leq n,\>1\leq k<l\leq n,\>1\leq m\leq n}$. Let $\lambda$ be
an indeterminate variable.

  1. _(1)_

Keep the notation of Lemma 5.2.1. The condition

(5.6) $x_{d}=\left(x_{i_{1}\ldots
i_{d}}\right)_{i_{1}<\ldots<i_{d}}\quad\text{is colinear to}\\\
(u+\lambda\operatorname{id})(x_{d})=\left(\sum_{j_{1}<\ldots<j_{d}}\Delta^{j_{1}\ldots
j_{d}}_{i_{1}\ldots i_{d}}(u+\lambda\operatorname{id})x_{j_{1}\ldots
j_{d}}\right)_{i_{1}<\ldots<i_{d}}$

translates into $d\binom{n}{d}\left(\binom{n}{d}-1\right)/2$ equations in $A$.

  2. _(2)_

The collection of all these equations in $A$ for $1\leq d\leq n-1$ defines
$\widetilde{X}$ inside $G/B\times\mathfrak{b}$.

###### Proof.

For all $i_{1}<\ldots<i_{d}$ and $j_{1}<\ldots<j_{d}$, the degree in $\lambda$
of $\Delta^{j_{1}\ldots j_{d}}_{i_{1}\ldots
i_{d}}(u+\lambda\operatorname{id})$ is at most $d-1$, unless
$(j_{1},\ldots,j_{d})=(i_{1},\ldots,i_{d})$ where the degree is $d$ with
leading term $\lambda^{d}$. Hence the degree of
$\sum_{j_{1}<\ldots<j_{d}}\Delta^{j_{1}\ldots j_{d}}_{i_{1}\ldots
i_{d}}(u+\lambda\operatorname{id})x_{j_{1}\ldots j_{d}}$ in $\lambda$ is $d$,
with leading term $x_{i_{1}\ldots i_{d}}\lambda^{d}$. Therefore the
$\binom{n}{d}\left(\binom{n}{d}-1\right)/2$ “cross-product” equations in
$A[\lambda]$ that express the condition (5.6) are of degree $d-1$ in
$\lambda$. This proves (1).

As for (2), the equations defining $\widetilde{X}$ are the ones that express
the condition $\operatorname{Ad}(g^{-1})\psi\in\mathfrak{b}$. They are
equivalent to saying that

$\operatorname{Ad}(g^{-1})(\psi+\lambda\operatorname{id})\in\mathfrak{b}$

for all $\lambda\in k$. Enlarging $k$ if necessary, we can assume that it is
algebraically closed, in particular infinite. Therefore there is a $\lambda\in
k$ such that $\psi+\lambda\operatorname{id}$ is invertible. Applying Lemma
5.2.1, our condition is equivalent to (5.6) for any such $\lambda$. Note that
(5.6) still holds true for any value of $\lambda$. Indeed, if
$(u+\lambda)(V_{d})$ is of dimension $<d$, then $(u+\lambda)(\sum
x_{i_{1}\ldots i_{d}}e_{i_{1}}\wedge\ldots\wedge e_{i_{d}})=0$, _i.e._ the
Plücker coordinates of $(u+\lambda)(V_{d})$ are zero. In particular their
vector is colinear to the vector of the Plücker coordinates of $V_{d}$.

In the end, $\operatorname{Ad}(g^{-1})\psi\in\mathfrak{b}$ if and only if
(5.6) is true for all $\lambda\in k$. As $k$ is algebraically closed, the
corresponding polynomials in $A[\lambda]$ are identically zero. ∎

######  Proposition 5.2.4.

Let $w\in\mathcal{S}_{n}$. Fix an integer $1\leq d\leq n-1$, let $A$ be the
ring $\mathbb{Z}[x_{i_{1}\ldots i_{d}},u_{kl},t_{m}]_{1\leq
i_{1}<\ldots<i_{d}\leq n,\>1\leq k<l\leq n,\>1\leq m\leq n}$, let $\lambda$ be
an indeterminate variable. For a sequence $1\leq i_{1}<\ldots<i_{d}\leq n$,
let $P_{w,i_{1}\ldots i_{d}}\in A[\lambda]$ be the polynomial

$P_{w,i_{1}\ldots
i_{d}}(\lambda)\coloneqq\sum_{j_{1}<\ldots<j_{d}}\Delta^{j_{1}\ldots
j_{d}}_{i_{1}\ldots i_{d}}(u+\lambda\operatorname{id})x_{j_{1}\ldots
j_{d}}-\left(\prod_{k=1}^{d}(t_{w(k)}+\lambda)\right)x_{i_{1}\ldots i_{d}}$

and for $0\leq s\leq d-1$, let $P_{w,i_{1}\ldots i_{d},s}\in A$ be the unique
homogeneous polynomials such that

$P_{w,i_{1}\ldots i_{d}}(\lambda)=\sum_{s=0}^{d-1}P_{w,i_{1}\ldots
i_{d},s}\lambda^{s}\,.$

Then, using the Plücker coordinates $(x_{i_{1}\ldots i_{d}})_{1\leq
d<n,\>1\leq i_{1}<\ldots<i_{d}\leq n}$ on $G/B$ and the affine coordinates
$(u_{kl},t_{m})_{1\leq k<l\leq n,\>1\leq m\leq n}$ on
$\mathfrak{b}=\mathfrak{u}\oplus\mathfrak{t}$, the subvariety
$\widetilde{V}_{w}$ of $G/B\times\mathfrak{b}$ is defined by:

  * •

the Plücker relations in dimension $d$, given in Proposition 5.1.5, for each
$1\leq d\leq n-1$

  * •

the incidence relations in dimension $(d,d+1)$, given in Proposition 5.1.6,
for each $1\leq d\leq n-2$

  * •

the equations for $(x_{i_{1}\ldots i_{d}})_{1\leq d<n,\>1\leq
i_{1}<\ldots<i_{d}\leq n}$ to be in the cell $BwB/B$ of $G/B$, given in
Proposition 5.1.8

  * •

the equations $P_{w,i_{1}\ldots i_{d},s}=0$ for all $1\leq d\leq n-1$, $1\leq
i_{1}<\ldots<i_{d}\leq n$ and $0\leq s\leq d-1$.

###### Proof.

Lemma 5.2.3 gives a desciption of $\widetilde{X}$ as a subvariety of
$G/B\times\mathfrak{b}$ defined by
$\sum_{d=1}^{n-1}d\binom{n}{d}\left(\binom{n}{d}-1\right)/2$ equations (which
are not necessarily independent). To find the equations for
$\widetilde{V}_{w}$, we add the ones given by Proposition 5.1.8.

Now, $x_{j_{1}\ldots j_{d}}=0$ for
$\\{j_{1},\ldots,j_{d}\\}\not\preceq\\{w(1),\ldots,w(d)\\}$ and
$\Delta^{j_{1}\ldots j_{d}}_{w(1)\ldots w(d)}(u+\lambda\operatorname{id})=0$
for $\\{j_{1},\ldots,j_{d}\\}\not\succeq\\{w(1),\ldots,w(d)\\}$. Hence the
$(w(1),\ldots,w(d))$-th term of the vector $(u+\lambda\operatorname{id})(x)$,
as expressed in (5.6), is

$\left(\prod_{k=1}^{d}(t_{w(k)}+\lambda)\right)x_{w(1)\ldots w(d)}\,.$

Knowing that $x_{w(1)\ldots w(d)}\neq 0$ and that $x_{j_{1}\ldots j_{d}}=0$
for $\\{j_{1},\ldots,j_{d}\\}\not\preceq\\{w(1),\ldots,w(d)\\}$, the
colinearity condition (5.6) is reduced to the following equations:

$\left(\prod_{k=1}^{d}(t_{w(k)}+\lambda)\right)x_{i_{1}\ldots
i_{d}}=\sum_{j_{1}<\ldots<j_{d}}\Delta^{j_{1}\ldots j_{d}}_{i_{1}\ldots
i_{d}}(u+\lambda\operatorname{id})x_{j_{1}\ldots j_{d}}$

where $i_{1}<\ldots<i_{d}$ is any strictly increasing sequence of indices.
These are the announced equations. ∎

######  Remark 5.2.5.

We have also seen that for each $w\in\mathcal{S}_{n}$, $1\leq d\leq n-1$ and
$1\leq i_{1}<\ldots<i_{d}\leq n$, the polynomial $P_{w,i_{1}\ldots i_{d}}$ is
a sum of terms

$a_{j_{1}\ldots j_{d}}x_{j_{1}\ldots
j_{d}},\quad\\{j_{1},\ldots,j_{d}\\}\succeq\\{i_{1},\ldots,i_{d}\\}\,,$

where $a_{j_{1}\ldots j_{d}}$ is an element of
$\mathbb{Z}[u_{kl},t_{m},\lambda]_{1\leq k<l\leq n,1\leq m\leq n}$. Also, the
particular “first term coefficient” $a_{i_{1}\ldots i_{d}}$ is equal to
$\prod_{k=1}^{d}(t_{i_{k}}+\lambda)-\prod_{k=1}^{d}(t_{w(k)}+\lambda)$. In
particular, for $\\{i_{1},\ldots,i_{d}\\}\succeq\\{w(1),\ldots,w(d)\\}$ the
polynomial $P_{w,i_{1}\ldots i_{d}}$ is in the ideal generated by the
$x_{j_{1}\ldots j_{d}}$ for
$\\{j_{1},\ldots,j_{d}\\}\succ\\{w(1),\ldots,w(d)\\}$, which are all zero on
the cell $BwB/B$ of $G/B$ by Proposition 5.1.8. Thus the associated equation
is tautological on $\widetilde{V}_{w}$.

###  5.3. Taking the Zariski closure

For $w\in\mathcal{S}_{n}$, let $\widetilde{X}^{\prime}_{w}$ denote the
subvariety of
$\prod_{d=1}^{n-1}\mathbb{P}(\bigwedge^{d}k^{n})\times\mathbb{A}^{n(n+1)/2}$,
with coordinate ring $A\coloneqq\mathbb{Z}[x_{j_{1}\ldots
j_{d}},u_{kl},t_{m}]$, defined by all the equations listed in Proposition
5.2.4 except for the inequations $x_{w(1)\ldots w(d)}\neq 0$ for $1\leq d\leq
n-1$. It is therefore the subvariety of $G/B\times\mathfrak{b}$ that satisfies
the equations $P_{w,i_{1}\ldots i_{d},s}=0$ for all $1\leq d\leq n-1$, $1\leq
i_{1}<\ldots<i_{d}\leq n$ and $0\leq s\leq d-1$. In particular, note that
$\widetilde{X}^{\prime}_{w}$ contains $\widetilde{V}_{w}$ by Proposition 5.2.4
and is Zariski-closed. Therefore
$\widetilde{X}_{w}\subseteq\widetilde{X}^{\prime}_{w}$.

######  Proposition 5.3.1.

Let $w,w^{\prime}\in\mathcal{S}_{n}$ such that $w^{\prime}\preceq w$. Then, as
algebraic varieties over $k$,

$\widetilde{X}^{\prime}_{w}\cap\widetilde{V}_{w^{\prime}}=\widetilde{V}_{w^{\prime}}\cap
q^{-1}(\mathfrak{t}^{ww^{\prime-1}}+\mathfrak{u})\,.$

###### Proof.

Let $(gB/B,\psi)\in G/B\times\mathfrak{b}$ be a point in the intersection
$\widetilde{X}^{\prime}_{w}\cap\widetilde{V}_{w^{\prime}}$. Its coordinates
$(x_{j_{1}\ldots j_{d}},u_{kl},t_{m})$ satisfy in particular $P_{w,i_{1}\ldots
i_{d}}(\lambda)=0$ in $k[\lambda]$ for any $\\{i_{1},\ldots,i_{d}\\}$ and
$x_{i_{1},\ldots,i_{d}}=0$ if
$\\{i_{1},\ldots,i_{d}\\}\not\preceq\\{w^{\prime}(1),\ldots,w^{\prime}(d)\\}$.
Under these latter conditions, almost all the terms in
$P_{w,w^{\prime}(1)\ldots w^{\prime}(d)}(\lambda)$ become zero, and the
equation $P_{w,w^{\prime}(1)\ldots w^{\prime}(d)}(\lambda)=0$ becomes

$\left(\prod_{k=1}^{d}(t_{w^{\prime}(k)}+\lambda)-\prod_{k=1}^{d}(t_{w(k)}+\lambda)\right)x_{w^{\prime}(1)\ldots
w^{\prime}(d)}=0$

on $\widetilde{X}^{\prime}_{w}$. Since we also have $x_{w^{\prime}(1)\ldots
w^{\prime}(d)}\neq 0$, we then must have
$\prod_{k=1}^{d}(t_{w^{\prime}(k)}+\lambda)=\prod_{k=1}^{d}(t_{w(k)}+\lambda)$
as polynomials in $k[\lambda]$.

Also, we have by definition, for all $\\{i_{1},\ldots,i_{d}\\}$,

$P_{w,i_{1}\ldots i_{d}}(\lambda)-P_{w^{\prime},i_{1}\ldots
i_{d}}(\lambda)=\prod_{k=1}^{d}(t_{w(k)}+\lambda)-\prod_{k=1}^{d}(t_{w^{\prime}(k)}+\lambda)$

in $A[\lambda]$. As a consequence, $P_{w,i_{1}\ldots i_{d},s}$ and
$P_{w^{\prime},i_{1}\ldots i_{d},s}$ are equal for all $s$ and
$\\{i_{1},\ldots,i_{d}\\}$ on $\widetilde{X}^{\prime}_{w}$. Therefore
$\widetilde{X}^{\prime}_{w}\cap\widetilde{V}_{w^{\prime}}$ is the subvariety
of $\widetilde{V}_{w^{\prime}}$ where
$\prod_{k=1}^{d}(t_{w^{\prime}(k)}+\lambda)=\prod_{k=1}^{d}(t_{w(k)}+\lambda)$
for all $1\leq d\leq n-1$. This last equality of polynomials in $k[\lambda]$
means that the multisets $\\{t_{w^{\prime}(1)},\ldots,t_{w^{\prime}(d)}\\}$
and $\\{t_{w(1)},\ldots,t_{w(d)}\\}$ are equal for all $1\leq d\leq n-1$. A
recursion on $d$ shows that this amounts to saying
$t_{w^{\prime}(k)}=t_{w(k)}$ for all $1\leq k\leq d$. For $d=n-1$, this is
equivalent to $(gB/B,\psi)\in
q^{-1}(\mathfrak{t}^{ww^{\prime-1}}+\mathfrak{u})$. ∎

Replacing $\widetilde{X}^{\prime}_{w}$ by $\widetilde{X}_{w}$ in Proposition
5.3.1 would yield Conjecture 3.2.3. However we shall see in Theorem 5.3.7 that
this is not always true.

The strategy to find counterexamples to Conjecture 3.2.3 is to use the
inequations $x_{w(1)\ldots w(d)}\neq 0$ satisfied on $\widetilde{V}_{w}$ to
simplify other equations satisfied on $\widetilde{V}_{w}$, by “removing” non-
zero factors. We then get an equation that is satisfied on
$\widetilde{V}_{w}$, hence on $\widetilde{X}_{w}$ as well, but not necessarily
on $\tilde{X}^{\prime}_{w}$. This is the object of the following two lemmas.
Once this is done, this equation, when considered on
$\widetilde{V}_{w^{\prime}}$, gives a condition that turns out to be strictly
more restrictive that simply being in
$q^{-1}(\mathfrak{t}^{ww^{\prime-1}}+\mathfrak{u})$.

######  Lemma 5.3.2.

Let $w\in\mathcal{S}_{n}$, $q\in\\{1,\ldots,n-2\\}$ and
$b\in\\{w(1),\ldots,w(q+1)\\}$; write
$\\{w(1),\ldots,w(q+1)\\}=\\{i_{1},\ldots,i_{q},b\\}$. Let
$\\{j_{1},\ldots,j_{q}\\}$ be a subset of $\\{1,\ldots,n\\}$ which does not
contain $b$. Assume that for any $1\leq l\leq q$ such that $j_{l}<b$, we have
$j_{l}\in\\{w(1),\ldots,w(q+1)\\}$. Then, for any $a\in\\{1,\ldots,n\\}$, the
equation

$x_{i_{1}\ldots i_{q}a}x_{j_{1}\ldots j_{q}b}=\pm x_{i_{1}\ldots
i_{q}b}x_{j_{1}\ldots j_{q}a}$

is satisfied on $\widetilde{V}_{w}$ for a certain choice of sign.

###### Proof.

Consider the Plücker relation in dimension $q+1$ with indices
$\\{i_{1},\ldots,i_{q}\\}$ and $\\{j_{1},\ldots,j_{q},a,b\\}$ (see Proposition
5.1.6), which is satisfied on $\widetilde{V}_{w}$. Up to signs, it is of the
form

(5.7) $\pm x_{i_{1}\ldots i_{q}a}x_{j_{1}\ldots j_{q}b}\pm x_{i_{1}\ldots
i_{q}b}x_{j_{1}\ldots j_{q}a}+\sum_{1\leq l\leq q}\pm x_{i_{1}\ldots
i_{q}k_{l}}x_{j_{1}\ldots\hat{j_{l}}\ldots j_{q}ab}=0\,.$

Let $1\leq l\leq q$. If $j_{l}<b$, then our hypothesis implies that
$x_{i_{1}\ldots i_{q}j_{l}}=0$ because the indices are redundant. Else we have
$j_{l}>b$, so that
$\\{i_{1},\ldots,i_{q},j_{l}\\}\succ\\{i_{1},\ldots,i_{q},b\\}=\\{w(1),\ldots,w(q+1)\\}$,
hence $x_{i_{1}\ldots i_{q}j_{l}}=0$ on $\widetilde{V}_{w}$ by virtue of
Proposition 5.1.8. Eliminating all those null variables in (5.7) gives the
desired equation. ∎

######  Lemma 5.3.3.

Keep the notation and assumption of Lemma 5.3.2. Assume further that
$\\{j_{1},\ldots,j_{q},a\\}\preceq\\{i_{1},\ldots,i_{q},b\\}$. Then the
equation

(5.8)
$\sum_{\\{j_{1},\ldots,j_{q}\\}\preceq\\{k_{1},\ldots,k_{q}\\}\preceq\\{i_{1},\ldots,i_{q}\\}}\pm\Delta^{k_{1}\ldots
k_{q}b}_{j_{1}\ldots j_{q}b}(u+\lambda\operatorname{id})x_{k_{1}\ldots
k_{q}a}-\left(\prod_{i=1}^{p+1}(t_{w(i)}+\lambda)\right)x_{j_{1}\ldots
j_{q}a}=0$

is satisfied on $\widetilde{V}_{w}$ for a certain choice of sign.

###### Proof.

We can assume that the sequences $(i_{l})_{1\leq l\leq q}$ and $(j_{l})_{1\leq
l\leq q}$ are in increasing order. Let $1\leq r\leq q$ be such that
$j_{r}<b<j_{r+1}$ (setting $j_{q+1}=\infty$ if necessary). Then
$\\{j_{1},\ldots,j_{q},a\\}$ has either $r$ or $r+1$ elements smaller than or
equal to $b$, depending on wether $a>b$ or $a\leq b$. Because
$\\{j_{1},\ldots,j_{q},a\\}\preceq\\{i_{1},\ldots,i_{q},b\\}$, this implies
that $\\{i_{1},\ldots,i_{q},b\\}$ has at most $r+1$ elements smaller than or
equal to $b$. However the hypothesis of Lemma 5.3.2 implies that
$\\{j_{1},\ldots,j_{r},b\\}\subseteq\\{i_{1},\ldots,i_{q},b\\}$. This forces
$j_{1}$,…,$j_{r}$ to be the $r$ smallest elements of
$\\{i_{1},\ldots,i_{q}\\}$, hence $i_{l}=j_{l}$ for all $l\leq r$.

In particular, we have
$\\{j_{1},\ldots,j_{q},b\\}\preceq\\{i_{1},\ldots,i_{q},b\\}$ and any set of
integers that sits between those two is of the form
$\\{k_{1},\ldots,k_{q},b\\}$ with $k_{l}=i_{l}$ for $l\leq r$ and $k_{l}>b$
for $l>r$. We can therefore apply Lemma 5.3.2 replacing
$\\{j_{1},\ldots,j_{q}\\}$ by $\\{k_{1},\ldots,k_{q}\\}$; this yields, up to
sign,

(5.9) $x_{i_{1}\ldots i_{q}a}x_{k_{1}\ldots k_{q}b}=\pm x_{i_{1}\ldots
i_{q}b}x_{k_{1}\ldots k_{q}a}\,.$

Now, keeping the notation of Proposition 5.2.4, we multiply by $x_{i_{1}\ldots
i_{q}a}$ the equation $P_{w,j_{1}\ldots j_{q}b}(\lambda)=0$ (in $A[\lambda]$).
From the discussion of Remark 5.2.5, the resulting equation only has terms in
$x_{k_{1}\ldots k_{q}b}x_{i_{1}\ldots i_{q}a}$ for
$\\{j_{1},\ldots,j_{q}\\}\preceq\\{k_{1},\ldots,k_{q}\\}$. Also, because of
Proposition 5.1.8, the $x_{k_{1}\ldots k_{q}b}$ are non-zero on
$\widetilde{V}_{w}$ only for
$\\{k_{1},\ldots,k_{q}\\}\preceq\\{i_{1},\ldots,i_{q}\\}$. With all these
simplifications, we get on $\widetilde{V}_{w}$ the equation

$\sum_{\\{j_{1},\ldots,j_{q}\\}\preceq\\{k_{1},\ldots,k_{q}\\}\preceq\\{i_{1},\ldots,i_{q}\\}}\Delta^{k_{1}\ldots
k_{q}b}_{j_{1}\ldots j_{q}b}(u+\lambda\operatorname{id})x_{k_{1}\ldots
k_{q}b}x_{i_{1}\ldots i_{q}a}\\\
-\left(\prod_{i=1}^{q+1}(t_{w(i)}+\lambda)\right)x_{j_{1}\ldots
j_{q}b}x_{i_{1}\ldots i_{q}a}=0\,.$

Using the equations (5.9), we get

$\sum_{\\{j_{1},\ldots,j_{q}\\}\preceq\\{k_{1},\ldots,k_{q}\\}\preceq\\{i_{1},\ldots,i_{q}\\}}\pm\Delta^{k_{1}\ldots
k_{q}b}_{j_{1}\ldots j_{q}b}(u+\lambda\operatorname{id})x_{k_{1}\ldots
k_{q}a}x_{i_{1}\ldots i_{q}b}\\\
-\left(\prod_{i=1}^{q+1}(t_{w(i)}+\lambda)\right)x_{j_{1}\ldots
j_{q}a}x_{i_{1}\ldots i_{q}b}=0$

which, after factorisation and simplification by $x_{i_{1}\ldots i_{q}b}$ (the
simplification is valid because $x_{i_{1}\ldots i_{q}b}=x_{w(1)\ldots
w(q+1)}\neq 0$ on $\widetilde{V}_{w}$), gives the equation we wanted. ∎

######  Proposition 5.3.4.

Let $w,w^{\prime}\in\mathcal{S}_{n}$ with $w^{\prime}\preceq w$. Assume that
there exist integers $a<b$ and $q\in\\{1,\ldots,n-2\\}$ such that:

  * •

$a\notin\\{w(1),\ldots,w(q+1)\\}$ and
$a\in\\{w^{\prime}(1),\ldots,w^{\prime}(q+1)\\}$,

  * •

$b\in\\{w(1),\ldots,w(q+1)\\}$ and
$b\notin\\{w^{\prime}(1),\ldots,w^{\prime}(q+1)\\}$,

  * •

for any $i\in\\{w^{\prime}(1),\ldots,w^{\prime}(q+1)\\}$ such that $i\neq a$
and $i<b$, we have $i\in\\{w(1),\ldots,w(q+1)\\}$.

Then the equation $t_{a}=t_{b}$ is satisfied on
$\widetilde{X}_{w}\cap\widetilde{V}_{w^{\prime}}$.

######  Example 5.3.5.

Take $n=4$, $w=[4231]$ and $w^{\prime}=[1324]$. Then $q=1$ and $(a,b)=(1,2)$
meets the condition of Proposition 5.3.4, hence
$\widetilde{X}_{w}\cap\widetilde{V}_{w^{\prime}}$ satisfies $t_{1}=t_{2}$.
This is stronger than for
$\widetilde{X}^{\prime}_{w}\cap\widetilde{V}_{w^{\prime}}$, which by
Proposition 5.3.1 only satisfies $t_{1}=t_{4}$ and $t_{2}=t_{3}$ (the
equations defining $\mathfrak{t}^{ww^{\prime-1}}$). One could also take
$(a,b)=(3,4)$; this gives the equation $t_{3}=t_{4}$ on
$\widetilde{X}_{w}\cap\widetilde{V}_{w^{\prime}}$, which is equivalent to the
already found $t_{1}=t_{2}$, since $t_{1}=t_{4}$ and $t_{2}=t_{3}$ are true.

###### Proof.

Write
$\\{j_{1},\ldots,j_{q}\\}=\\{w^{\prime}(1),\ldots,w^{\prime}(q+1)\\}\setminus\\{a\\}$;
then the hypotheses of Lemma 5.3.2 are satisfied. From Example 2.4.4 applied
to $w^{\prime}\preceq w$, we also have
$\\{j_{1},\ldots,j_{q},a\\}\preceq\\{i_{1},\ldots,i_{q},b\\}$. We can
therefore apply Lemma 5.3.3 to get

$\sum_{\\{j_{1},\ldots,j_{q}\\}\preceq\\{k_{1},\ldots,k_{q}\\}\preceq\\{i_{1},\ldots,i_{q}\\}}\pm\Delta^{k_{1}\ldots
k_{q}b}_{j_{1}\ldots j_{q}b}(u+\lambda\operatorname{id})x_{k_{1}\ldots
k_{q}a}-\left(\prod_{i=1}^{p+1}(t_{w(i)}+\lambda)\right)x_{j_{1}\ldots
j_{q}a}=0$

which is satisfied on $\widetilde{V}_{w}$, hence on the Zariski closure
$\widetilde{X}_{w}$ as well.

Now consider this equation on
$\widetilde{X}_{w}\cap\widetilde{V}_{w^{\prime}}$. From Proposition 5.1.8, all
variables $x_{k_{1}\ldots k_{q}a}$ are zero on $\widetilde{V}_{w^{\prime}}$
for $\\{k_{1},\ldots,k_{q}\\}\succ\\{j_{1},\ldots,j_{q}\\}$, so it reduces to
one term:

$\left(\Delta^{j_{1}\ldots j_{q}b}_{j_{1}\ldots
j_{q}b}(u+\lambda\operatorname{id})-\prod_{i=1}^{p+1}(t_{w(i)}+\lambda)\right)x_{j_{1}\ldots
j_{q}a}=0\,.$

Again on $\widetilde{V}_{w^{\prime}}$ we can simplify by $x_{j_{1}\ldots
j_{q}a}=x_{w^{\prime}(1)\ldots w^{\prime}(q+1)}\neq 0$. Moreover, $u$ being
upper triangular gives $\Delta^{j_{1}\ldots j_{q}b}_{j_{1}\ldots
j_{q}b}(u+\lambda\operatorname{id})=(t_{b}+\lambda)\prod_{l=1}^{p}(t_{j_{l}}+\lambda)=\frac{t_{b}+\lambda}{t_{a}+\lambda}\prod_{i=1}^{p+1}(t_{w^{\prime}(i)}+\lambda)$,
so the equation becomes

$\frac{t_{b}+\lambda}{t_{a}+\lambda}\prod_{i=1}^{p+1}(t_{w^{\prime}(i)}+\lambda)=\prod_{i=1}^{p+1}(t_{w(i)}+\lambda)\,.$

We also know that
$\prod_{i=1}^{p+1}(t_{w^{\prime}(i)}+\lambda)=\prod_{i=1}^{p+1}(t_{w(i)}+\lambda)$
from the proof of Proposition 5.3.1; so we finally get
$t_{a}+\lambda=t_{b}+\lambda$, which implies $t_{a}=t_{b}$. ∎

######  Remark 5.3.6.

The same result holds when the last condition on $a,b$ in Proposition 5.3.4 is
replaced by the following variant: for any
$j\in\\{w(1),\ldots,w(q+1)\\}\setminus\\{b\\}$ such that $j>a$, we have
$j\in\\{w^{\prime}(1),\ldots,w^{\prime}(q+1)\\}$. The proof carries out in
almost identical fashion, with the difference that, in the proof of Lemma
5.3.2, instead of using the Plücker relation with indices
$\\{i_{1},\ldots,i_{q}\\}$ and $\\{k_{1},\ldots,k_{q},a,b\\}$ for equation
(5.7), we use the Plücker relation with indices $\\{k_{1},\ldots,k_{q}\\}$ and
$\\{i_{1},\ldots,i_{q},a,b\\}$.

######  Theorem 5.3.7.

Let $w,w^{\prime}\in\mathcal{S}$ with $w^{\prime}\preceq w$. Assume that there
exist $a$, $b$ as in Proposition 5.3.4 such that $a$ and $b$ are not in the
same orbit under $ww^{\prime-1}$. Then Conjecture 3.2.1 is false for the pair
$(w,w^{\prime})$, that is:

$\widetilde{V}_{w^{\prime}}\cap
q^{-1}(\mathfrak{t}^{ww^{\prime-1}}+\mathfrak{u})\not\subseteq\widetilde{X}_{w}\,.$

###### Proof.

Consider any point $x=(x_{i_{1}\ldots
i_{d}},t_{i},u_{ij})\in\prod_{d=1}^{n-1}\mathbb{P}(\bigwedge^{d}k^{n})\times\mathbb{A}^{n(n+1)/2}$
where, for $1\leq d\leq n-1$ and $1\leq i_{1}<\ldots<i_{d}<n$, $x_{i_{1}\ldots
i_{d}}\neq 0$ if and only if
$\\{i_{1},\ldots,i_{d}\\}=\\{w^{\prime}(1),\ldots,w^{\prime}(d)\\}$, and where
$u_{ij}=0$ for all $1\leq i<j\leq n$. It is easily checked that
$x\in\widetilde{V}_{w^{\prime}}$, either from the equations of Proposition
5.2.4, or from the fact that such a point $x$ corresponds to $(gB/B,\psi)\in
G/B\times\mathfrak{b}$ where $g$ is an invertible matrix with zeroes
everywhere but in positions $(w^{\prime}(j),j)$ for $1\leq j\leq n$, and where
$\psi\in\mathfrak{t}$.

When $a$ and $b$ are not in the same orbit under $ww^{\prime-1}$, it is
possible to set $\psi=(t_{1},\ldots,t_{n})\in\mathfrak{t}^{ww^{\prime-1}}$
with $t_{a}\neq t_{b}$. This gives $x\in
q^{-1}(\mathfrak{t}^{ww^{\prime-1}}+\mathfrak{u})$, while Proposition 5.3.4
forces $x\notin\widetilde{X}_{w}\cap\widetilde{V}_{w^{\prime}}$, so
$x\notin\widetilde{X}_{w}$. ∎

######  Remark 5.3.8.

In the situation of Theorem 5.3.7, we actually have a strict inclusion

$\widetilde{X}_{w}\cap\widetilde{V}_{w^{\prime}}\subsetneq\widetilde{V}_{w^{\prime}}\cap
q^{-1}(\mathfrak{t}^{ww^{\prime-1}}+\mathfrak{u})$

since the (non strict) inclusion is already known (see Proposition 5.3.1 or
the proof of Proposition 3.2.4). It is also known that
$\widetilde{V}_{w^{\prime}}\cap
q^{-1}(\mathfrak{t}^{ww^{\prime-1}}+\mathfrak{u})$ is irreducible: by an
argument similar to the proof of Proposition 3.2.4, this statement is
equivalent to the analogous statement with $V_{w^{\prime}}$ (defined in §3.1)
in place of $\widetilde{V}_{w^{\prime}}$, which is [8, Lemma 2.3.5]. Therefore
$\widetilde{X}_{w}\cap\widetilde{V}_{w^{\prime}}$ is a union of irreducible
component which have dimension strictly smaller than the dimension of
$\widetilde{V}_{w^{\prime}}\cap
q^{-1}(\mathfrak{t}^{ww^{\prime-1}}+\mathfrak{u})$, thus

$\dim\left(\widetilde{X}_{w}\cap\widetilde{V}_{w^{\prime}}\right)<\dim\left(\widetilde{V}_{w^{\prime}}\cap
q^{-1}(\mathfrak{t}^{ww^{\prime-1}}+\mathfrak{u})\right)\,.$

######  Remark 5.3.9.

The easiest example of $w,w^{\prime}$ meeting the conditions of Theorem 5.3.7
is when there exist $1\leq i<j<k<l\leq n$ and $1\leq a<b<c<d\leq n$ such that:

  1. (1)

$(w(i),w(j),w(k),w(l))=(d,b,c,a)$.

  2. (2)

$(w^{\prime}(i),w^{\prime}(j),w^{\prime}(k),w^{\prime}(l))=(a,c,b,d)$.

  3. (3)

for all $m\in\\{1,\ldots,n\\}\setminus\\{i,j,k,l\\}$, $w(m)=w^{\prime}(m)$.

The smallest such example is when $(w,w^{\prime})$ is the first bad pair for
$n=4$ (see Theorem 2.4.10). In fact, all cases of interest (_e.g._ when
$(w,w^{\prime})$ is a bad pair, see Lemma 5.3.10 below) for $n\leq 5$ meet the
conditions of Theorem 5.3.7.

For $n=6$, a few cases arise where neither those conditions nor the variant of
Remark 5.3.6 are met, for example when $w=(16)(25)$ and $w^{\prime}=(34)$
(using the cycles notation for elements of the symmetric group
$\mathcal{S}_{6}$). In this specific case, the approach of the proof of
Theorem 5.3.7 is not adapted. Indeed, the only equations defining the Schubert
cell $BwB/B$ in $G/B$ are $x_{6}\neq 0$, $x_{56}\neq 0$, $x_{356}\neq 0$,
$x_{3456}\neq 0$, $x_{23456}\neq 0$ and $x_{456}=0$. Because $x_{456}$ is the
only zero variable, there are too few Plücker relations and incidence
relations that have only two nonzero terms on $BwB/B$ (in the same way
equation (5.7) gives Lemma 5.3.2). Furthermore those few relations cannot be
used to get new equations satisfied on $\widetilde{V}_{w}$ (and hence
$\widetilde{X}_{w}$) that are nontrivial on $\widetilde{V}_{w^{\prime}}$ (the
same way the equation $t_{a}=t_{b}$ of Proposition 5.3.4 is), because all
Plücker coordinates but $x_{1},x_{12},x_{123},x_{124},x_{1234},x_{12345}$ are
zero on $\widetilde{V}_{w^{\prime}}$. The status of Conjecture 3.2.1 remains
unknown for this pair.

Finally, the following lemma confirms that Theorem 5.3.7 and Theorem 4.1 are
consistent with each other.

######  Lemma 5.3.10.

Let $(w^{\prime},w)\in W$ satisfy the conditions of Theorem 5.3.7. Then it is
a bad pair in the sense of §2. The same conclusion holds with the conditions
of Remark 5.3.6.

###### Proof.

If the pair $(w,w^{\prime})$ satisfies the conditions of Theorem 5.3.7,
including the one that states that $a$ and $b$ are not in the same orbit under
$ww^{\prime-1}$, then it is a bad pair in the sense of §2. Indeed, let
$\Omega$ the orbit of $b$ under $ww^{\prime-1}$. By assumption,
$a\notin\Omega$, so the last condition of Proposition 5.3.4 implies that

$\left(\Omega\cap
w(\\{1,\ldots,q+1\\})\right)\cap\\{1,\ldots,b\\}\supsetneq\left(\Omega\cap
w^{\prime}(\\{1,\ldots,q+1\\})\right)\cap\\{1,\ldots,b\\}$

with the inequality stemming from the presence of $b$ in the left-hand side
and not in the right-hand side. Therefore $\Omega\cap
w(\\{1,\ldots,q+1\\})\succeq\Omega\cap w^{\prime}(\\{1,\ldots,q+1\\})$ is
false (see the second characterisation in Definition 2.4.2), so the pair
$(w,w^{\prime})$ is bad according to Proposition 2.4.5.

In the situation of Remark 5.3.6, take $\Omega$ to be the orbit of $a$ under
$ww^{\prime-1}$. Again, $b\notin\Omega$, so the condition spelled out in
Remark 5.3.6 implies that

$\left(\Omega\cap
w(\\{1,\ldots,q+1\\})\right)\cap\\{a,\ldots,n\\}\subsetneq\left(\Omega\cap
w^{\prime}(\\{1,\ldots,q+1\\})\right)\cap\\{a,\ldots,n\\}$

from which we get $\lvert(\Omega\cap
w(\\{1,\ldots,q+1\\}))\cap\\{a,\ldots,n\\}\rvert<\lvert(\Omega\cap
w^{\prime}(\\{1,\ldots,q+1\\}))\cap\\{a,\ldots,n\\}\rvert$, or equivalently
$\lvert(\Omega\cap
w(\\{1,\ldots,q+1\\}))\cap\\{1,\ldots,a-1\\}\rvert>\lvert(\Omega\cap
w^{\prime}(\\{1,\ldots,q+1\\}))\cap\\{1,\ldots,a-1\\}\rvert$. As before, this
means that $\Omega\cap w(\\{1,\ldots,q+1\\})\succeq\Omega\cap
w^{\prime}(\\{1,\ldots,q+1\\})$ is false and hence $(w,w^{\prime})$ is a bad
pair. ∎

##  Appendix A Jordan decomposition and orbits in $\mathfrak{g}$

We recall here a few well-known results about Jordan decompositions in Lie
algebras. Note that Proposition A.9 below is needed in §4.2. We thank Anne
Moreau for its proof, which we couldn’t find precisely in the literature.

In the following, $k$ is an algbraically closed field of characteristic $0$
(so that $k$-points and closed points on algebraic varieties over $k$ are the
same). For instance, there is no distinction between viewing a Lie algebra as
a $k$-scheme (as we do in §4) or as a $k$-vector space.

######  Definition A.1.

Let $G$ be an algebraic group over $k$ with Lie algebra $\mathfrak{g}$. A
closed point $x\in\mathfrak{g}$ is called _semisimple_ (resp.  _nilpotent_) if
the following property holds. For any morphism of algebraic groups $\phi\colon
G\longrightarrow\operatorname{GL}(V)$ inducing
$\mathop{}\mathopen{}\mathrm{d}\phi\colon\mathfrak{g}\longrightarrow\operatorname{\mathfrak{gl}}(V)$,
where $V$ is a finite-dimensional vector space over $k$, the endomorphism
$\mathop{}\mathopen{}\mathrm{d}\phi(x)\in\operatorname{\mathfrak{gl}}(V)\simeq\operatorname{End}(V)$
is diagonalisable (resp. nilpotent).

######  Theorem A.2 (Jordan decomposition).

  1. _(1)_

Let $G$ be an algebraic group over $k$ with Lie algebra $\mathfrak{g}$. For
any closed point $x\in\mathfrak{g}$, there is a unique couple
$(x_{\mathrm{s}},x_{\mathrm{n}})\in\mathfrak{g}^{2}$ with $x_{\mathrm{s}}$
semisimple and $x_{\mathrm{n}}$ nilpotent such that
$[x_{\mathrm{s}},x_{\mathrm{n}}]=0$ and $x=x_{\mathrm{s}}+x_{\mathrm{n}}$. We
call this the _Jordan decomposition_ of $x$.

  2. _(2)_

The Jordan decomposition is functorial: for any morphism $\phi\colon
G\longrightarrow H$ of algebraic groups over $k$ inducing
$\mathop{}\mathopen{}\mathrm{d}\phi\colon\mathfrak{g}\longrightarrow\mathfrak{h}$
and any closed point $x\in\mathfrak{g}$, we have

$\mathop{}\mathopen{}\mathrm{d}\phi(x_{\mathrm{s}})=(\mathop{}\mathopen{}\mathrm{d}\phi(x))_{\mathrm{s}},\quad\mathop{}\mathopen{}\mathrm{d}\phi(x_{\mathrm{n}})=(\mathop{}\mathopen{}\mathrm{d}\phi(x))_{\mathrm{n}}\,.$

###### Proof.

The existence of a functorial Jordan decomposition is a combination of 23.6.2,
23.6.3 and 23.6.4 of [26]. The uniqueness property in (1) is given by the fact
that for any algebraic group $G$, there exists a finite-dimensional vector
space $V$ with a morphism of algebraic groups $\phi\colon
G\longrightarrow\operatorname{GL}(V)$ such that
$\mathop{}\mathopen{}\mathrm{d}\phi\colon\mathfrak{g}\longrightarrow\operatorname{\mathfrak{gl}}(V)$
is injective (see 22.1.5 and 24.4.1 of _loc. cit._). ∎

######  Proposition A.3.

Let $G$ be a connected reductive group over $k$, let $\mathfrak{g}$ be the Lie
algebra of $G$ and $\mathfrak{z}$ be the centre of $\mathfrak{g}$. There is a
decomposition of $k$-vector spaces

(A.1) $\mathfrak{g}=[\mathfrak{g},\mathfrak{g}]\oplus\mathfrak{z}$

and for any closed point $x\in\mathfrak{g}$, the following conditions are
equivalent:

  1. _(i)_

$x$ is semisimple (resp. nilpotent).

  2. _(ii)_

$\operatorname{ad}(x)\in\operatorname{\mathfrak{gl}}(\mathfrak{g})$ is
semisimple (resp.
$\operatorname{ad}(x)\in\operatorname{\mathfrak{gl}}(\mathfrak{g})$ is
nilpotent and the component of $x$ in $\mathfrak{z}$ in (A.1) is $0$).

  3. _(iii)_

For any finite-dimensional representation
$\sigma\colon\mathfrak{g}\longrightarrow\operatorname{\mathfrak{gl}}(V)$, the
endomorphism
$\sigma(x)\in\operatorname{\mathfrak{gl}}(V)\simeq\operatorname{End}(V)$ is
diagonalisable (resp. nilpotent).

###### Proof.

The decomposition (A.1) is 27.2.2 and 20.5.5 of [26]. Note that
$\mathfrak{z}=0$ if and only if $G$ is semisimple (see 20.1.2 and 20.5.4 of
_loc. cit._). The implication (iii)$\implies$(i) is a consequence of the
definitions. We now prove (i)$\implies$(ii). Since
$\operatorname{ad}\colon\mathfrak{g}\longrightarrow\operatorname{\mathfrak{gl}}(\mathfrak{g})$
is the differential of $\operatorname{Ad}\colon
G\longrightarrow\operatorname{GL}(\mathfrak{g})$ (see 23.5.5 of _loc. cit._),
we only need to prove the following statement (the rest follows from the
definitions): if $x\in\mathfrak{g}$ decomposes as $x=y+z$ via (A.1) and $x$ is
nilpotent, then $z=0$. Since $[\mathfrak{g},\mathfrak{g}]$ is the Lie algebra
of the derived group $(G,G)$ (see 27.2.3 of _loc. cit._), we can consider the
Jordan decomposition $y=y_{\mathrm{s}}+y_{\mathrm{n}}$ of $y$ in
$[\mathfrak{g},\mathfrak{g}]$. Applying Theorem A.2(2) to the inclusion
$(G,G)\lhook\joinrel\relbar\joinrel\rightarrow G$, we see that
$y_{\mathrm{s}}$ (resp. $y_{\mathrm{n}}$) is semisimple (resp. nilpotent) in
$\mathfrak{g}$. As $z$ is semisimple in $\mathfrak{g}$ (see 27.2.2(ii) of
_loc. cit._) and commutes with everything, we see that $y_{\mathrm{s}}+z$ is
semisimple and $x=(y_{\mathrm{s}}+z)+y_{\mathrm{n}}$ is the Jordan
decomposition of $x$. By uniqueness, $y_{\mathrm{s}}+z=0$ hence $z=0$. This
finishes the proof of (i)$\implies$(ii). Finally, (ii)$\implies$(iii) follows
from 20.5.7 and 20.5.8 of _loc. cit._ (note that their definition of
semisimple and nilpotent elements differs from ours). ∎

######  Lemma A.4.

Let $G$ be a semisimple algebraic group over $k$, with semisimple Lie algebra
$\mathfrak{g}$. Let $\mathfrak{h},\mathfrak{h}^{\prime}$ be Cartan subalgebras
of $\mathfrak{g}$. Then there exists a closed point $g\in G$ such that
$\mathfrak{h}^{\prime}=\operatorname{Ad}_{G}(g)\mathfrak{h}$.

###### Proof.

We proceed exactly as in [26, 29.2.3(ii)]. We first recall the notation of
19.1.1 of _loc. cit_. For a closed point $x\in\mathfrak{g}$, we write

$\mathfrak{g}^{0}(x)\coloneqq\cap_{n\in\mathbb{Z}_{\geq
0}}\ker((\operatorname{ad}{x})^{n})$

and we say that $x$ is _generic_ if $\dim{\mathfrak{g}^{0}(x)}$ (which is the
multiplicity of $0$ as a root of the characteristic polynomial of
$\operatorname{ad}(x)$) is minimal among all $x\in\mathfrak{g}$. Call
$\mathfrak{g}_{\mathrm{gen}}$ the set of generic elements; it is a Zariski-
open subset of $\mathfrak{g}$ since the coefficients of the characteristic
polynomial are a polynomial expression in $x\in\mathfrak{g}$. By 29.2.3(i) of
_loc. cit._ a Lie subalgebra of $\mathfrak{g}$ is Cartan if and only if it is
of the form $\mathfrak{g}^{0}(x)$ for some $x\in\mathfrak{g}_{\mathrm{gen}}$.
Hence choose $x\in\mathfrak{g}_{\mathrm{gen}}$ such that
$\mathfrak{h}=\mathfrak{g}^{0}(x)$. Consider the morphism

$\theta\colon\;\begin{aligned}
G\times\mathfrak{h}\;&\longrightarrow\;\mathfrak{g}\\\
(g,y)\;&\longmapsto\;\operatorname{Ad}(g)y\end{aligned}$

whose differential at the point $(e,x)\in G\times\mathfrak{h}$ ($e$ is the
neutral element in $G$) is

$\mathop{}\mathopen{}\mathrm{d}\theta_{(e,x)}\colon\;\begin{aligned}
\mathfrak{g}\times\mathfrak{h}\;&\longrightarrow\;\mathfrak{g}\\\
(z,y)\;&\longmapsto\;[z,x]+y\end{aligned}$

(see 29.1.4(i) and 23.5.5 of _loc. cit._). By definition,
$\mathfrak{h}=\mathfrak{g}^{0}(x)$ is a subspace of $\mathfrak{g}$ stable by
$\operatorname{ad}(x)$, and $\operatorname{ad}(x)$ induces an automorphism on
$\mathfrak{g}/\mathfrak{g}^{0}(x)=\mathfrak{g}/\mathfrak{h}$. Therefore, as in
29.2.1 of _loc. cit._ , $\mathop{}\mathopen{}\mathrm{d}\theta_{(e,x)}$ is
surjective, so $\theta$ is dominant by 16.5.7 of _loc. cit._ , and from 15.4.2
of _loc. cit._ , the image $\operatorname{Ad}(G)\mathfrak{h}$ contains an open
dense subset of $\mathfrak{g}$. Since the same can be said of
$\operatorname{Ad}(G)\mathfrak{h}^{\prime}$, we have in $\mathfrak{g}$

$\operatorname{Ad}(G)\mathfrak{h}\cap\operatorname{Ad}(G)\mathfrak{h}^{\prime}\cap\mathfrak{g}_{\mathrm{gen}}\neq\emptyset\,.$

Therefore there is $t\in\mathfrak{g}_{\mathrm{gen}}$ and $g,g^{\prime}\in G$
such that $\operatorname{Ad}(g)t\in\mathfrak{h}$ and
$\operatorname{Ad}(g^{\prime})t\in\mathfrak{h}^{\prime}$. Since
$\operatorname{Ad}(g)t$ and $\operatorname{Ad}(g^{\prime})t$ are generic as
well, 19.8.5 of _loc. cit._ gives

$\mathfrak{h}=\mathfrak{g}^{0}(\operatorname{Ad}(g)t)\,,\quad\mathfrak{h}^{\prime}=\mathfrak{g}^{0}(\operatorname{Ad}(g^{\prime})t)$

hence $\mathfrak{h}^{\prime}=\operatorname{Ad}(g^{\prime}g^{-1})\mathfrak{h}$.
∎

For any vector space $V$ over any field, write $V^{*}$ for its dual, $S(V)$
for the symmetric algebra of $V$ and $S^{n}(V)$ for the graded piece of $S(V)$
of degree $n$ (where $n\in\mathbb{Z}_{\geq 0}$). Recall that if $V$ is finite
dimensional, $S(V^{*})$ can be interpreted as the algebra of polynomial
functions on $V$.

Let $\mathfrak{g}$ be a semisimple Lie algebra, the action of $\mathfrak{g}$
on itself by adjunction induces an action on $S(\mathfrak{g}^{*})$. We can
then consider the algebra $S(\mathfrak{g}^{*})^{\mathfrak{g}}$ of invariants
under this action. Similarly, given a Cartan subalgebra $\mathfrak{h}$ of
$\mathfrak{g}$, the Weyl group $W$ of $(\mathfrak{g},\mathfrak{h})$ acts on
$\mathfrak{h}$ (see [26, §19.8]), so we can consider $S(\mathfrak{h}^{*})^{W}$
as well.

######  Theorem A.5.

Let $\mathfrak{g}$ be a semisimple Lie algebra over $k$, $\mathfrak{h}$ be a
Cartan subalgebra of $\mathfrak{g}$ and $W$ be the Weyl group of the root
system associated to $(\mathfrak{g},\mathfrak{h})$.

  1. _(1)_

The restriction morphism $S(\mathfrak{g}^{*})\longrightarrow
S(\mathfrak{h}^{*})$ induces a surjection
$S(\mathfrak{g}^{*})^{\mathfrak{g}}\relbar\joinrel\twoheadrightarrow
S(\mathfrak{h}^{*})^{W}$.

  2. _(2)_

For $n\in\mathbb{Z}_{\geq 0}$, the vector space $S^{n}(\mathfrak{g}^{*})$ is
spanned by functions of the form
$x\longmapsto\operatorname{tr}(\sigma(x)^{n})$, where $\sigma$ is a finite-
dimensional representation of $\mathfrak{g}$ over $k$.

###### Proof.

This is [26, 31.2.6]. ∎

######  Lemma A.6.

Let $\mathfrak{g}$ be a semisimple Lie algebra over $k$, $\mathfrak{h}$ be a
Cartan subalgebra of $\mathfrak{g}$ and $W$ be the Weyl group of the root
system associated with $(\mathfrak{g},\mathfrak{h})$. Let $x,y\in\mathfrak{h}$
be closed points such that $f(x)=f(y)$ for all $f\in S(\mathfrak{h}^{*})^{W}$.
Then $x$ and $y$ are in the same orbit for the action of $W$ on
$\mathfrak{h}$.

###### Proof.

This is [26, 34.2.1]. ∎

######  Proposition A.7.

Let $G$ be a semisimple algebraic group over $k$ with Lie algebra
$\mathfrak{g}$, let $x,y\in\mathfrak{g}$ be semisimple elements. Assume that
$\operatorname{tr}(\sigma(x)^{n})=\operatorname{tr}(\sigma(y)^{n})$ for all
finite-dimensional representation $\sigma$ of $\mathfrak{g}$ and all
$n\in\mathbb{Z}_{\geq 0}$. Then $x$ and $y$ are in the same orbit for the
action of $G$ on $\mathfrak{g}$ by adjunction.

###### Proof.

We know from [26, 20.6.3(ii)] that $x$ and $y$ are each contained in a Cartan
subalgebra of $\mathfrak{g}$. According to Lemma A.4, we can assume that $x$
and $y$ lie in the same Cartan subalgebra $\mathfrak{h}$. Let $W$ be the Weyl
group associated to the root system of $(\mathfrak{g},\mathfrak{h})$. It
follows from Theorem A.5 and Lemma A.6 that $x$ and $y$ are in the same orbit
for the action of $W$ on $\mathfrak{h}$.

There is a maximal torus $T$ of $G$ such that $\mathfrak{h}$ is the Lie
algebra of $T$ (see 29.2.5 of _loc. cit._ and [14, 26.2.A]). The action of the
Weyl group $W(G,T)=N_{G}(T)/T$ on $T$ induces, through differentiation, an
action of $\mathfrak{h}$; this identifies $W(G,T)$ with
$W=W(\mathfrak{g},\mathfrak{h})$. Hence $W$ is isomorphic to $N_{G}(T)/T$, and
the corresponding action of $w\in N_{G}(T)/T$ on $\mathfrak{h}$ is given by
the adjunction by a lifting of $w$ in $N_{G}(T)$. Therefore there is an
element $n\in N_{G}(T)\subset G$ such that $x=\operatorname{Ad}_{G}(n)y$. ∎

######  Lemma A.8.

Let $\mathfrak{g}$ be a semisimple Lie algebra over $k$, let
$\sigma\colon\mathfrak{g}\longrightarrow\operatorname{\mathfrak{gl}}(V)$ be a
finite-dimensional representation and let $n\in\mathbb{Z}_{\geq 0}$.

  1. _(1)_

Let $x\in\mathfrak{g}$ be a closed point with Jordan decomposition
$x=x_{\mathrm{s}}+x_{\mathrm{n}}$. Then
$\operatorname{tr}(\sigma(x)^{n})=\operatorname{tr}(\sigma(x_{\mathrm{s}})^{n})$.

  2. _(2)_

Let $\mathfrak{b}$ be a Borel subalgebra of $\mathfrak{g}$, let
$x,t\in\mathfrak{b}$ be closed points such that $x-t$ is nilpotent. Then
$\operatorname{tr}(\sigma(x)^{n})=\operatorname{tr}(\sigma(t)^{n})$.

###### Proof.

The assertion (1) is [26, 34.2.3]. As for (2), note that by semisimplicity
$\mathfrak{g}=[\mathfrak{g},\mathfrak{g}]$, so $\sigma(\mathfrak{b})$ is a
solvable subalgebra of
$\operatorname{\mathfrak{sl}}(V)=[\operatorname{\mathfrak{gl}}(V),\operatorname{\mathfrak{gl}}(V)]$.
Hence $\sigma(\mathfrak{b})$ is contained in a Borel subalgebra
$\mathfrak{b}^{\prime}$ of $\operatorname{\mathfrak{sl}}(V)$. Using 29.4.7 of
_loc. cit._ , one can find a basis of $V$ such that $\mathfrak{b}^{\prime}$ is
the set of upper triangular matrices in $\operatorname{\mathfrak{sl}}(V)$.
Since $x-t$ is nilpotent, by Proposition A.3 the matrix $\sigma(x)-\sigma(t)$
is nilpotent and in $\mathfrak{b}^{\prime}$ hence is strictly upper
triangular. Therefore $\sigma(x)$ and $\sigma(t)$ are upper triangular with
the same diagonal in $\mathfrak{b}^{\prime}$. Thus so are $\sigma(x)^{n}$ and
$\sigma(t)^{n}$, which therefore have the same trace on $V$. ∎

######  Proposition A.9.

Let $G$ be a connected reductive group over $k$ with a maximal torus $T$ and a
Borel subgroup $B$ containing $T$, let $\mathfrak{g}$, $\mathfrak{t}$ and
$\mathfrak{b}$ be their respective Lie algebra. Let $x\in\mathfrak{b}$ be a
closed point with Jordan decomposition $x=x_{\mathrm{s}}+x_{\mathrm{n}}$.

  1. _(1)_

The components $x_{\mathrm{s}}$ and $x_{\mathrm{n}}$ are in $\mathfrak{b}$.

  2. _(2)_

Let $t\in\mathfrak{t}$ and $u\in\mathfrak{b}$ such that $u$ is nilpotent and
$x=t+u\ \in\mathfrak{b}$. Then there exists $g\in(G,G)$ such that
$x_{\mathrm{s}}=\operatorname{Ad}_{G}(g)t$.

###### Proof.

The inclusion $\mathfrak{b}\subset\mathfrak{g}$ is the differential of the
inclusion $B\subset G$, hence by Theorem A.2(2), the Jordan decomposition of
$x$ viewed as an element of $\mathfrak{b}$ is the same as that of $x$ viewed
as an element of $\mathfrak{g}$. This proves (1).

For (2), the decomposition (A.1) gives $x=y+z$ with
$y\in[\mathfrak{g},\mathfrak{g}]$ and $z\in\mathfrak{z}$. Proposition A.3
gives $u\in[\mathfrak{g},\mathfrak{g}]$, hence $t=s+z$ for some semisimple
element $s\in[\mathfrak{g},\mathfrak{g}]$. Since $\mathfrak{z}$ is the Lie
algebra of the center $Z(G)$ of $G$ (see [26, 24.4.2(ii)]) and $Z(G)\subset
T\subset B$ (see [14, 26.2.A]), we have
$\mathfrak{z}\subset\mathfrak{t}\subset\mathfrak{b}$, hence
$y=x-z\in\mathfrak{b}$ and $s=t-z\in\mathfrak{b}$, thus
$y,s\in\mathfrak{b}^{\prime}\coloneqq\mathfrak{b}\cap[\mathfrak{g},\mathfrak{g}]$.

Note that $\mathfrak{b}=\mathfrak{b}^{\prime}\oplus\mathfrak{z}$ since
$\mathfrak{z}\subset\mathfrak{b}$. Let $\mathfrak{b}_{1}^{\prime}$ be a
solvable Lie subalgebra of $[\mathfrak{g},\mathfrak{g}]$ containing
$\mathfrak{b}^{\prime}$ and
$\mathfrak{b}_{1}\coloneqq\mathfrak{b}_{1}^{\prime}\oplus\mathfrak{z}\subseteq\mathfrak{g}$.
Then $\mathfrak{b}_{1}$ is a Lie subalgebra of $\mathfrak{g}$ containing
$\mathfrak{b}$ and it is solvable since
$[\mathfrak{b}_{1},\mathfrak{b}_{1}]=[\mathfrak{b}_{1}^{\prime},\mathfrak{b}_{1}^{\prime}]$.
Therefore $\mathfrak{b}_{1}=\mathfrak{b}$, and
$\mathfrak{b}_{1}^{\prime}=\mathfrak{b}^{\prime}$. This shows that
$\mathfrak{b}^{\prime}$ is a Borel subalgebra of
$[\mathfrak{g},\mathfrak{g}]$. Consequently, since $y=x-z=t+u-z=s+u$, we can
apply Lemma A.8 and Proposition A.7 to the semisimple group $(G,G)$ and the
elements $y$, $s$ of its Lie algebra $[\mathfrak{g},\mathfrak{g}]$ (see [26,
27.2.3]). This gives $g\in(G,G)\subset G$ such that

$y_{\mathrm{s}}=\operatorname{Ad}_{(G,G)}(g)s=\operatorname{Ad}_{G}(g)s\,.$

From the proof of Proposition A.3, we know that
$x_{\mathrm{s}}=y_{\mathrm{s}}+z$. Since the conjugation by $g$ is the
identity on $Z(G)$, its differential $\operatorname{Ad}_{G}(g)$ is trivial on
$\mathfrak{z}$, so

$x_{\mathrm{s}}=y_{\mathrm{s}}+z=\operatorname{Ad}_{G}(g)(s+z)=\operatorname{Ad}_{G}(g)t$

which finishes the proof. ∎

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