On the index of the quotient of a Borel subalgebra by an ad-nilpotent ideal

TL;DR

This paper investigates bounds for the index of quotients of Borel subalgebras by ad-nilpotent ideals in simple Lie algebras, generalizing known formulas and proposing conjectures for exactness.

Contribution

It provides new upper bounds for these indices and extends Panov's formula from type A to more general cases, including conjecturing exactness.

Findings

Upper bounds for the index of Borel quotient are established.

The bound for the nilpotent radical quotient generalizes Panov's type A formula.

The bound for the Borel quotient is exact in type A and conjectured to be in general.

Abstract

In this paper, we give upper bounds for the index of the quotient of the Borel subalgebra of a simple Lie algebra or its nilpotent radical by an ad-nilpotent ideal. For the nilpotent radical quotient, our bound is a generalization of the formula for the index given by Panov in the type A case. In general, this bound is not exact. Using results from Panov, we show that the upper bound for the Borel quotient is exact in the type $A$ case, and we conjecture that it is exact in general.

Equations

$$\chi (\mathfrak{a})=\underset{f\in {\mathfrak{a}}^{\ast }}{\min }\dim {\mathfrak{a}}^f$$

$${\mathfrak{g}}^P=\underset{\alpha \in P}{⨁}{\mathfrak{g}}^{\alpha }.$$

$$\begin{array}[]{rcl}\chi(\mathfrak{a},V)&=&\dim V-\max\limits_{h\in V^{*}}\dim\mathfrak{a}.h=\dim V-\max\limits_{h\in V^{*}}\operatorname{codim}_{\mathfrak{a}}\mathfrak{a}^{h}\\ &=&\mathrm{tr\,deg}_{\Bbbk}(\Bbbk(V^{*})^{A})\end{array}$$

$$H(E,\gamma )=\{\alpha \in E;\gamma -\alpha \in E\cup \{0\}\}.$$

$$E_1=E\setminus H(E,{\theta }_1),{\Gamma }_1=H(E,{\theta }_1).$$

$$E_{i+1}=E_i\setminus H(E_i,{\theta }_{i+1}),{\Gamma }_{i+1}=H(E_i,{\theta }_{i+1}).$$

$$\begin{array}[]{ll}\Gamma_{1}=\{\alpha_{1,3},\alpha_{1,1},\alpha_{2,3},\alpha_{1,2},\alpha_{3,3}\},&\Gamma_{4}=\{\alpha_{4,6},\alpha_{4,5},\alpha_{6,6}\},\\ \Gamma_{2}=\{\alpha_{2,5},\alpha_{2,2},\alpha_{3,5},\alpha_{2,4},\alpha_{5,5}\},&\Gamma_{5}=\{\alpha_{4,4}\},\\ \Gamma_{3}=\{\alpha_{3,6},\alpha_{3,4},\alpha_{5,6}\}.\\ \end{array}$$

$${\Phi }_f(X,Y)=f([X,Y]),$$

$$\chi (\mathfrak{a})=\min \{\mathrm{corank}{\Phi }_f;f\in {\mathfrak{a}}^{\ast }\}.$$

$${\mathfrak{q}}_{\Phi }\simeq \mathfrak{h}\oplus {\mathfrak{g}}^{{\Delta }^{+}\setminus \Phi },{\mathfrak{m}}_{\Phi }\simeq {\mathfrak{g}}^{{\Delta }^{+}\setminus \Phi }.$$

$$f_{\mathbf{a}}=\sum _{i=1}^s{a}_i{X}_{{\theta }_i}^{\ast }.$$

$${\mathcal{G}}_i=\{(\alpha ,\beta )\in {\Gamma }_i\times {\Gamma }_i;\alpha +\beta ={\theta }_i\text{mbox}and\alpha <\beta \},t_i=♯{\mathcal{G}}_i,$$

$$\mathcal{G}=\bigcup _{i=1}^s{\mathcal{G}}_i,t=♯\mathcal{G}.$$

$$v_z=X_{\alpha }^{\ast }\wedge X_{\beta }^{\ast }\in {\bigwedge }^2{\mathfrak{q}}_{\Phi }^{\ast }.$$

$${\Phi }_{f_{\mathbf{a}}}={\Psi }_{f_{\mathbf{a}}}+{\Theta }_{f_{\mathbf{a}}}$$

$${\Theta }_{f_{\mathbf{a}}}=\sum _{i=1}^s{K}_i\wedge X_{{\theta }_i}^{\ast },{\Psi }_{f_{\mathbf{a}}}=\sum _{z\in \mathcal{Z}}{\lambda }_z{v}_z,$$

$${\lambda }_z={\Phi }_{f_{\mathbf{a}}}(X_{\alpha },X_{\beta })=f_{\mathbf{a}}([X_{\alpha },X_{\beta }])={\mu }_z{a}_i.$$

$$\dim \mathfrak{t}=\dim \mathfrak{h}-\dim D(\mathbf{h})=\ell -d(\mathbf{h}).$$

$$f_{\mathbf{a}}([h_{k},X_{\theta_{j}}])=\left\{\begin{array}[]{ll}a_{i_{k}}&\mbox{if }j=i_{k},\\ 0&\mbox{otherwise}.\\ \end{array}\right.$$

$${\Theta }_{f_{\mathbf{a}}}=\sum _{i\in \mathcal{I}}{a}_i{\theta }_i\wedge X_{{\theta }_i}^{\ast }+\sum _{j\notin \mathcal{I}}{K}_j\wedge X_{{\theta }_j}^{\ast },$$

$${\bigwedge }^t{\Psi }_{f_{\mathbf{a}}}=t!\sum _{1⩽i_1<\cdots <i_t⩽n}{\lambda }_{i_1}\cdots {\lambda }_{i_t}{v}_{i_1}\wedge \cdots \wedge v_{i_t}.$$

$$\operatornamewithlimits{\textstyle\prod}_{i=1}^{s}a_{i}^{t_{i}}\Big{(}\operatornamewithlimits{\textstyle\prod}_{z\in{\mathcal{G}}}\mu_{z}\Big{)}.$$

$$\left({\Gamma }_j\setminus \{{\theta }_j\}\right)\cap \left(\bigcup _{k\in \mathcal{I}}\widetilde{z_{i_k}}\right)\ne \varnothing .$$

$${\mu }_{i_1}\cdots {\mu }_{i_t}\prod _{i=1}^s{a}_i^{m_i},$$

$${\bigwedge }^{r+t}{\Phi }_{f_{\mathbf{a}}}=\sum _{k=0}^{r+t}\left(\begin{array}{c}r+t\\ k\end{array}\right)\left({\bigwedge }^k{\Psi }_{f_{\mathbf{a}}}\right)\wedge \left({\bigwedge }^{r+t-k}{\Theta }_{f_{\mathbf{a}}}\right).$$

$${\bigwedge }^t{\Psi }_{f_{\mathbf{a}}}=\lambda v_1\wedge \cdots \wedge v_t+w,$$

$$\begin{array}[]{l}\chi(\mathfrak{q}_{\Phi})\leqslant\mathrm{min}\{\ell+\ell(\mathbf{h})-2d(\mathbf{h});\mathbf{h}\in{\mathcal{H}}\},\\ \chi(\mathfrak{m}_{\Phi})\leqslant\mathrm{min}\{\ell(\mathbf{h});\mathbf{h}\in{\mathcal{H}}\}.\end{array}$$

$$\begin{array}[]{l}\dim\mathfrak{q}_{\Phi}=\dim\mathfrak{h}+\ell(\mathbf{h})+2t,\\ \dim\mathfrak{m}_{\Phi}=\ell(\mathbf{h})+2t.\end{array}$$

$$\dim {\mathfrak{q}}_{\Phi }^{f_{\mathbf{a}}}⩽\dim {\mathfrak{q}}_{\Phi }-2(d(\mathbf{h})+t).$$

$$\dim {\mathfrak{q}}_{\Phi }^{f_{\mathbf{a}}}⩽\dim \mathfrak{h}+\ell (\mathbf{h})-2d(\mathbf{h}).$$

Full text

Résumé. Dans cet article, nous donnons des bornes supérieures pour l’indice du quotient d’une sous-algèbre de Borel d’une algèbre de Lie simple ou de son radical nilpotent par un idéal ad-nilpotent. Pour le quotient du radical nilpotent, notre borne supérieure est une généralisation de la formule obtenue pour l’indice par Panov en type $A$. En général, cette borne n’est pas exacte. En utilisant des résultats de Panov, nous montrons qu’en ce qui concerne le quotient de la sous-algèbre de Borel, notre borne supérieure est exacte dans le cas du type $A$, et nous conjecturons que c’est aussi le cas en général.

1. Introduction

The index of the quotient of a Borel subalgebra by an ad-nilpotent ideal is considered in recent works of P. Damianou, H. Sabourin and P. Vanhaecke [2] on problems related to Toda-lattices and integrable systems. They associate to such an ad-nilpotent ideal an Hamiltonian system, and they want to determine whether this system is integrable.

Let $\mathfrak{g}$ be a simple finite-dimensional Lie algebra over an algebraically closed field $\Bbbk$ of characteristic zero, and $\mathfrak{b}$ a Borel subalgebra of $\mathfrak{g}$. For any ad-nilpotent ideal $\mathfrak{i}$ of $\mathfrak{b}$, $(\mathfrak{b}/\mathfrak{i})^{*}$ is a Poisson submanifold of $\mathfrak{b}^{*}$. Its Poisson rank $L$ is equal to the dimension of $\mathfrak{b}/\mathfrak{i}$ minus the index of $\mathfrak{b}/\mathfrak{i}$. Since the number of equations required for the previous Hamiltonian system to be integrable is $\dim(\mathfrak{b}/\mathfrak{i})^{*}-L/2$, the calculation of the index of $\mathfrak{b}/\mathfrak{i}$ is involved in this problem.

Recall that the index of a finite-dimensional Lie algebra $\mathfrak{a}$ over $\Bbbk$ is the integer

[TABLE]

where for $f\in\mathfrak{g}^{*}$, we denote by $\mathfrak{a}^{f}=\{X\in\mathfrak{a};f([X,Y])=0$ for all $Y\in\mathfrak{a}\}$, the annihilator of $f$ for the coadjoint representation of $\mathfrak{a}$. It is well-known that when $\mathfrak{a}$ is the Lie algebra of an algebraic group $A$, $\chi(\mathfrak{a})$ is the transcendence degree of the field of $A$-invariant rational functions on $\mathfrak{a}^{*}$.

There are quite a lot of recent works on the computation of the index of certain classes of Lie subalgebras of a semisimple Lie algebra : parabolic subalgebras and related subalgebras ([3], [4], [8], [11], [6]), centralizers of elements and related subalgebras ([9], [1], [13], [5]).

Let $\mathfrak{h}$ be a Cartan subalgebra of the simple Lie algebra $\mathfrak{g}$ contained in $\mathfrak{b}$, $\Delta$ the associated root system, $\Delta^{+}$ the set of positive roots relative to $\mathfrak{b}$ and $\Pi=\{\alpha_{1},\dots,\alpha_{\ell}\}$ the corresponding set of simple roots. For each $\alpha\in\Delta$, let $\mathfrak{g}^{\alpha}$ be the root subspace of $\mathfrak{g}$ relative to $\alpha$. Denote by $\mathfrak{n}=\mathfrak{g}^{\Delta^{+}}$ the nilpotent radical of $\mathfrak{b}$ where for a subset $P$ of $\Delta^{+}$, we set

[TABLE]

An ideal $\mathfrak{i}$ of $\mathfrak{b}$ is ad-nilpotent if and only if for all $x\in\mathfrak{i}$, $\mathrm{ad}_{\mathfrak{b}}x$ is nilpotent. Since any ideal of $\mathfrak{b}$ is $\mathfrak{h}$-stable, we deduce easily that an ideal is ad-nilpotent if and only if it is nilpotent, and there exists a subset $\Phi\subset\Delta^{+}$ such that $\mathfrak{i}=\mathfrak{g}^{\Phi}$. We set $\mathfrak{q}_{\Phi}=\mathfrak{b}/\mathfrak{i}$ and $\mathfrak{m}_{\Phi}=\mathfrak{n}/\mathfrak{i}$.

In [7], Panov determined the index of $\mathfrak{m}_{\Phi}$, when $\mathfrak{g}$ is simple of type $A$. His results are very explicit, and the index is completely determined by $\Delta^{+}\setminus\Phi$ in a combinatorial way. A similar consideration of roots was used for the index of seaweed subalgebras in [11]. In this paper, we generalize these root combinatorial approaches to give upper bounds for the index of $\mathfrak{q}_{\Phi}$ and $\mathfrak{m}_{\Phi}$ in all types. Our upper bound for $\mathfrak{m}_{\Phi}$ is not exact when $\mathfrak{g}$ is not of type $A$. However, using the results of Panov for $\mathfrak{m}_{\Phi}$, we prove that our upper bound for $\mathfrak{q}_{\Phi}$ is exact when $\mathfrak{g}$ is of type $A$, and we have not found so far any counter-examples in the other types. We give also a short discussion on the existence of stable linear forms.

We shall recall a more general definition of the index which is used in the paper. Let $\mathfrak{a}$ be the Lie algebra of an algebraic group $A$ and $V$ a rational $A$-module of finite dimension. The index of $V$ is the integer

[TABLE]

where for $f\in V^{*}$, $\mathfrak{a}^{f}=\{X\in\mathfrak{a};X.f=0\}$ and $\mathfrak{a}.f=\{X.f;X\in\mathfrak{a}\}$. When $f\in V^{*}$ is such that $\dim V-\dim\mathfrak{a}.f=\chi(\mathfrak{a},V),$ we say that $f$ is regular. The set of regular elements of $V^{*}$ is a non-empty Zariski-open subset.

2. H-sequences

In this section, we introduce the combinatorial tools used to describe the upper bounds for the index of the quotients. This is a generalization of the “cascade” construction of Kostant (see for example [11, 12]) and the construction of Panov in type $A$ in [7].

Recall the following standard partial order on $\Delta^{+}$. For $\alpha,\beta\in\Delta^{+}$, we have $\alpha\leqslant\beta$ if and only if $\beta-\alpha$ is a sum of positive roots. Let $E\subset\Delta^{+}$ and $\gamma\in E$. We set :

[TABLE]

Definition 2.1**.**

Let $E\subset\Delta^{+}$ and $\theta_{1}\in E$. We say that $(\theta_{1})$ is an H-sequence of length $1$ in $E$ if $E=H(E,\theta_{1})$.

By induction, for $\theta_{1},\theta_{2},\dots,\theta_{r}\in E$, we say that $(\theta_{1},\theta_{2},\dots,\theta_{r})$ is an H-sequence of length $r$ in $E$ if and only if :

1. (i)

$(\theta_{2},\theta_{3},\dots,\theta_{r})$* is an H-sequence of length $r-1$ in $E\setminus H(E,\theta_{1})$,* 2. (ii)

$\theta_{1}$* is a maximal element for $\leqslant$ in $E$.*

Let $(\theta_{1},\theta_{2},\dots,\theta_{r})$ be an H-sequence of length $r$ in $E$. Set :

[TABLE]

For $i=1,\dots,r-1$, we set

[TABLE]

It is clear from the definition that $E$ is the disjoint union of $\Gamma_{1},\dots,\Gamma_{r}$, and we have $E_{i}=E_{i+1}\cup\Gamma_{i+1}$ for $i=0,\dots,r-1$, with the convention that $E_{0}=E$.

Let $\mathbf{h}$ be an H-sequence. We denote by $\ell(\mathbf{h})$ its length, $D(\mathbf{h})$ the vector space in $\mathfrak{h}^{*}$ spanned by the elements in $\mathbf{h}$, and $d(\mathbf{h})=\dim D(\mathbf{h})$.

Examples 2.2**.**

(i) Let $E=\Delta^{+}$. We recover (an ordered) Kostant’s cascade construction of pairwise strongly orthogonal roots in $\Delta^{+}$.

(ii) Let $\mathfrak{g}$ be of type $A_{6}$. Using the numbering of simple roots in [12], set $\alpha_{i,j}=\alpha_{i}+\dots+\alpha_{j}$. Take $\Phi=\{\alpha\in\Delta^{+};\alpha\geqslant\alpha_{1,4}\mbox{ or }\alpha\geqslant\alpha_{2,6}\}$, and $E=\Delta^{+}\setminus\Phi$. Then $\mathbf{h}=\{\alpha_{1,3},\alpha_{2,5},\alpha_{3,6},\alpha_{4,6},\alpha_{4,4}\}$ is an H-sequence of length $5$ in $E$, where

[TABLE]

We have another H-sequence $\mathbf{h}^{\prime}=(\alpha_{2,5},\alpha_{3,6},\alpha_{4,4},\alpha_{6,6},\alpha_{1,3},\alpha_{1,2},\alpha_{1,1}).$ which is of length $7$. Observe that $d(\mathbf{h})=5$ and $d(\mathbf{h}^{\prime})=6$.

Lemma 2.3**.**

Let $E\subset\Delta^{+}$ and $\mathbf{h}=(\theta_{1},\dots,\theta_{r})$ be an H-sequence of length $r$ in $E$. Let $i,j,k\in\{1,\dots,r\}$.

1. (i)

Let $\alpha\in\Gamma_{i}$ and $\beta\in\Gamma_{j}$ be such that $\alpha+\beta=\theta_{k}$. Then $k\geqslant\min(i,j)$. 2. (ii)

There do not exist $i,j,k$ such that $\theta_{i}+\theta_{j}=\theta_{k}$.

Proof.

(i) If $k<\min(i,j)$, then $\alpha,\beta\in E_{k}$. It follows that $\alpha,\beta\in\Gamma_{k}$ and $k=j=i$, which contradicts the hypothesis.

(ii) Assume that there exist $i,j,k$ such that $\theta_{i}+\theta_{j}=\theta_{k}$. Then $\theta_{k}>\theta_{i}$ and $\theta_{k}>\theta_{j}$ and therefore by construction $k<\min(i,j)$, which contradicts the first point. ∎

3. Upper bounds for the index

We give in this section upper bounds for the index of the quotients. The proof follows closely to the one for the index of seaweed Lie algebras in [11] even though we do not have the nice properties on the roots from the “cascade construction”.

Recall that if $\mathfrak{a}$ is a finite-dimensional Lie algebra over $\Bbbk$ and $f\in\mathfrak{a}^{*}$, we can define an alternating bilinear form $\Phi_{f}$ on $\mathfrak{a}$ by setting

[TABLE]

for $X,Y\in\mathfrak{a}$. Then $\mathfrak{a}^{f}=\{X\in\mathfrak{a};\Phi_{f}(X,Y)=0,\mbox{ for all }Y\in\mathfrak{a}\}$ is the kernel of $\Phi_{f}$. Therefore we have

[TABLE]

Let $\{H_{1},\dots,H_{\ell}\}$ be a basis of $\mathfrak{h}$. For $\alpha\in\Delta$, we denote by $X_{\alpha}$ a non-zero element of $\mathfrak{g}^{\alpha}$. Then $\{H_{i};1\leqslant i\leqslant\ell\}\cup\{X_{\alpha};\alpha\in\Delta\}$ is a basis of $\mathfrak{g}$ and we shall denote by $\{H_{i}^{*};1\leqslant i\leqslant\ell\}\cup\{X_{\alpha}^{*};\alpha\in\Delta\}$ the corresponding dual basis.

Let $\Phi$ be a subset of $\Delta^{+}$ such that $\mathfrak{i}=\mathfrak{g}^{\Phi}$ is an ad-nilpotent ideal of $\mathfrak{b}$. Suppose that $\mathbf{h}=(\theta_{1},\dots,\theta_{s})$ is an H-sequence of length $s$ of $E=\Delta^{+}\setminus\Phi$. We have the following $\mathfrak{h}$-module isomorphisms :

[TABLE]

Let $\mathbf{a}=(a_{1},\dots,a_{s})$ be an element of $(\Bbbk^{*})^{s}$. Identifying $\mathfrak{q}_{\Phi}^{*}$ with $\mathfrak{h}^{*}\oplus\displaystyle\sum_{\alpha\in\Delta^{+}\setminus\Phi}\Bbbk X_{\alpha}^{*}$, we define the following element of $\mathfrak{q}_{\Phi}^{*}$ :

[TABLE]

We fix a total order $<$ on $\Delta^{+}$ compatible with the partial order $\leqslant$.

For $i\in\{1,\dots,s\}$, set

[TABLE]

and

[TABLE]

Denote by ${\mathcal{Z}}$ the set of pairs $(\alpha,\beta)$ of $E^{2}$ such that $\alpha<\beta$ and there exists $k\in\{1,\dots,s\}$ satisfying $\alpha+\beta=\theta_{k}$.

For $z=(\alpha,\beta)\in{\mathcal{Z}}$, we set

[TABLE]

Identifying $\Phi_{f_{\mathbf{a}}}$ with an element of $\bigwedge^{2}\mathfrak{q}_{\Phi}^{*}$, we have

[TABLE]

where

[TABLE]

with $K_{i}\in D(\mathbf{h})$ for $i=1,\dots,s$ and $\lambda_{z}\in\Bbbk$ for all $z\in{\mathcal{Z}}$.

If $z=(\alpha,\beta)\in\mathcal{Z}$, then $\Theta_{f_{\mathbf{a}}}(X_{\alpha},X_{\beta})=0$. Moreover, we have $[X_{\alpha},X_{\beta}]=\mu_{z}X_{\theta_{i}}$, for some $i\in\{1,\dots,s\}$ and a non-zero scalar $\mu_{z}$. Consequently,

[TABLE]

Thus $\lambda_{z}$ is non-zero.

Lemma 3.1**.**

In the above notations :

1. (i)

$\mathfrak{q}_{\Phi}^{f_{\mathbf{a}}}$* contains a commutative subalgebra of $\mathfrak{q}_{\Phi}$, consisting of semi-simple elements and of dimension $\ell-d(\mathbf{h}).$* 2. (ii)

We have $\bigwedge^{d(\mathbf{h})}\Theta_{f_{\mathbf{a}}}\neq 0$ and $\bigwedge^{d(\mathbf{h})+1}\Theta_{f_{\mathbf{a}}}=0$. 3. (iii)

There exists a non-empty open subset $U$ of $(\Bbbk^{*})^{s}$ such that we have $\bigwedge^{t}\Psi_{f_{\mathbf{a}}}\neq 0$ and $\bigwedge^{d(\mathbf{h})+t}\Phi_{f_{\mathbf{a}}}\neq 0$ whenever $\mathbf{a}\in U$.

Proof.

The proof is similar to the one for the Lemme in [11, §3.9].

(i) For simplicity, we write $\mathfrak{q}=\mathfrak{q}_{\Phi}$. Let $\mathfrak{t}=\{x\in\mathfrak{h};\theta_{i}(x)=0\mbox{ for }i=1,\dots,s\}$ be the orthogonal of $D(\mathbf{h})$ in $\mathfrak{h}$. Then :

[TABLE]

We also have that $\displaystyle[\mathfrak{t},\mathfrak{q}]\subset\bigoplus_{\alpha\in E\setminus\{\theta_{1},\dots,\theta_{s}\}}\mathfrak{g}^{\alpha}$. It follows that $\mathfrak{t}$ is contained in $\mathfrak{q}^{f_{\mathbf{a}}}$, and therefore, we obtain the result.

(ii) Set $r=d(\mathbf{h})$. Let ${\mathcal{I}}=\{i_{1},\dots,i_{r}\}\subset\{1,\dots,s\}$ be such that $(\theta_{i_{1}},\dots,\theta_{i_{r}})$ is a basis of $D(\mathbf{h})$ and complete to a basis $\mathcal{B}^{\prime}=(\beta_{1},\dots,\beta_{\ell})$ of $\mathfrak{h}^{*}$ such that $\beta_{k}=\theta_{i_{k}}$ for $k=1,\dots,r$. Denote by $\mathcal{B}=(h_{1},\dots,h_{\ell})$ the basis of $\mathfrak{h}$ dual to $\mathcal{B}^{\prime}$. Then we have,

[TABLE]

It follows that

[TABLE]

where $K_{j}\in D(\mathbf{h})$. The result follows easily beacause $\sharp{\mathcal{I}}=d(\mathbf{h})$.

(iii) If $z,z^{\prime}\in\mathcal{Z}$, we have $v_{z}\wedge v_{z^{\prime}}=v_{z^{\prime}}\wedge v_{z}$ and $v_{z}\wedge v_{z}=0$. Let $z_{1},\dots,z_{n}$ be the elements of $\mathcal{Z}$ such that $z_{1},\dots,z_{t}$ are the elements of ${\mathcal{G}}$. For simplicity, let us write $\lambda_{i}v_{i}$ for $\lambda_{z_{i}}v_{z_{i}}$ and $\mu_{i}$ for $\mu_{z_{i}}$. Consequently:

[TABLE]

In the previous sum, the coefficient of $v_{1}\wedge\cdots\wedge v_{t}$ is by (1)

[TABLE]

Now assume that $v_{i_{1}}\wedge\cdots\wedge v_{i_{t}}=\lambda v_{1}\wedge\cdots\wedge v_{t}$, with $\lambda\in\Bbbk^{*}$, where $i_{1}<\cdots<i_{t}$ and $(i_{1},\dots,i_{t})\neq(1,\dots,t)$.

If $z=(\alpha,\beta)\in{\mathcal{Z}}$, we denote by $\widetilde{z}=\{\alpha,\beta\}$ the underlying set of $z$. Then the set $\mathcal{S}=\widetilde{z}_{1}\cup\cdots\cup\widetilde{z_{t}}$ is the disjoint union of the sets $\widetilde{z_{i_{k}}}$ for $1\leqslant k\leqslant t$. It follows that if $z_{i_{k}}\notin{\mathcal{G}}$, then we have by Lemma 2.3 that there exist $i,j\in\{1,\dots,s\}$ such that $i\not=j$ and $z_{i_{k}}=(\alpha,\beta)$ where $(\alpha,\beta)\in(\Gamma_{i}\setminus\{\theta_{i}\})\times(\Gamma_{j}\setminus\{\theta_{j}\})$.

Let ${\mathcal{I}}=\{k;z_{i_{k}}\not\in{\mathcal{G}}\}$. Let $i_{0}$ be minimal among the elements $j\in\{1,\dots,s\}$ verifying :

[TABLE]

Then there exists $\alpha\in\Gamma_{i_{0}}$, $k\in{\mathcal{I}}$ and $\beta\in\Delta^{+}$ such that $\widetilde{z_{i_{k}}}=\{\alpha,\beta\}$. By our choice of $i_{0}$ and since $z_{i_{k}}\not\in{\mathcal{G}}$, there exists $j\geqslant i_{0}$ and $l\in\{1,\dots,s\}$ such that $\beta\in\Gamma_{j}$ and $\alpha+\beta=\theta_{l}$. Then, by Lemma 2.3, we have $l>\min(i_{0},j)=i_{0}$. It follows that $\lambda_{z_{i_{k}}}=\mu_{z_{i_{k}}}a_{l}$, where $l\neq i_{0}$.

We deduce that the coefficient of $v_{i_{1}}\wedge\cdots\wedge v_{i_{t}}$ in the sum giving $\bigwedge^{t}\Psi_{f_{\mathbf{a}}}$ is of the form

[TABLE]

with $m_{i_{0}}<t_{i_{0}}$.

It is now clear that there exists a non-empty open subset $U$ of $(\Bbbk^{*})^{s}$ verifying $\bigwedge^{t}\Psi_{f_{\mathbf{a}}}\neq 0$ if $\mathbf{a}\in U$.

Finally, we have

[TABLE]

Set $\displaystyle\mathfrak{u}=\sum_{i=1}^{s}\Bbbk X_{\theta_{i}}$. Since $\bigwedge^{j}\Theta_{f_{\mathbf{a}}}\in(\textstyle{\bigwedge^{j}}D(\mathbf{h}))\wedge(\textstyle{\bigwedge^{j}}\mathfrak{u}^{*})$, to show that $\bigwedge^{r+t}\Phi_{f_{\mathbf{a}}}\neq 0$, it suffices to prove that $(\bigwedge^{t}\Psi_{f_{\mathbf{a}}})\wedge(\bigwedge^{r}\Theta_{f_{\mathbf{a}}})\neq 0$.

If $\mathbf{a}\in U$, then we deduce from the preceding paragraphes that

[TABLE]

where $\lambda\in\Bbbk^{*}$ and $w$ is a linear combination of elements of the form $v_{z_{i_{1}}}\wedge\cdots\wedge v_{z_{i_{t}}}$, with $\widetilde{z_{i_{1}}}\cup\cdots\cup\widetilde{z_{i_{t}}}\neq\mathcal{S}$. It is therefore clear that $(\bigwedge^{t}\Psi_{f_{\mathbf{a}}})\wedge(\bigwedge^{r}\Theta_{f_{\mathbf{a}}})\neq 0$ if $\mathbf{a}\in U$. ∎

Theorem 3.2**.**

Let $\Phi$ be a subset of $\Delta^{+}$ such that $\mathfrak{g}^{\Phi}$ is an ad-nilpotent ideal of $\mathfrak{g}$. Denote by ${\mathcal{H}}$ the set of H-sequences of $\Delta^{+}\setminus\Phi$. Then, we have

[TABLE]

Proof.

Let $\mathbf{h}\in{\mathcal{H}}$ with $\ell(\mathbf{h})=s$ and $t$ as defined in the beginning of this section. By definition, we have

[TABLE]

Let $U$ be a non-empty open subset of $(\Bbbk^{*})^{s}$ verifying part (iii) of Lemma 3.1. If $\mathbf{a}\in U$, then the fact that $\bigwedge^{d(\mathbf{h})+t}\Phi_{f_{\mathbf{a}}}\neq 0$ implies that $\operatorname{rk}(\Phi_{f_{\mathbf{a}}})\geqslant 2(d(\mathbf{h})+t)$. Thus

[TABLE]

Hence

[TABLE]

In the same manner, if $\mathbf{a}\in U$, then the fact that $\bigwedge^{t}\Psi_{f_{\mathbf{a}}}\neq 0$ implies that $\operatorname{rk}(\Psi_{f_{\mathbf{a}}})\geqslant 2t$. Thus

[TABLE]

So we are done. ∎

For an H-sequence $\mathbf{h}$, we define

[TABLE]

Proposition 3.3**.**

Let us conserve the notations of Theorem 3.2. If $\mathbf{h}\in{\mathcal{H}}$ verifies $c(\mathbf{h})\in\{0,1\}$, then $\chi(\mathfrak{q}_{\Phi})=c(\mathbf{h})$.

Proof.

The case $c(\mathbf{h})=0$ is clear by Theorem 3.2. So let us suppose that $c(\mathbf{h})=1$.

We have $\dim\mathfrak{q}_{\Phi}-c(\mathbf{h})=2(d(\mathbf{h})+t)$. Since $\dim\mathfrak{q}_{\Phi}-\chi(\mathfrak{q}_{\Phi})$ is an even integer (it is the rank of an alternating bilinear form on $\mathfrak{q}_{\Phi}$), we deduce that $c(\mathbf{h})$ and $\chi(\mathfrak{q}_{\Phi})$ are of the same parity. So $\chi(\mathfrak{q}_{\Phi})=1$. ∎

4. Type A

Let us assume in this section that $\mathfrak{g}$ is of type $A_{\ell}$. We shall show that the upper bound for the index of $\mathfrak{q}_{\Phi}$ is exact in this special case.

We fix a subset $\Phi$ of $\Delta^{+}$ such that $\mathfrak{i}=\mathfrak{g}^{\Phi}$ is an ad-nilpotent ideal of $\mathfrak{g}$. As in example 2.2, we use the numbering of simple roots in [12], and we set $\alpha_{i,j}=\alpha_{i}+\dots+\alpha_{j}$ when $i\leqslant j$.

We fix the following total order $\prec$ on $\Delta^{+}$ compatible with the partial order $\leqslant$ :

[TABLE]

It is clear that there is a unique H-sequence $\mathbf{h}=(\theta_{1},\dots,\theta_{s})$ of $E=\Delta^{+}\setminus\Phi$, satisfying $\theta_{1}\succ\theta_{2}\succ\dots\succ\theta_{s}$. This H-sequence is considered by Panov in [7], and we shall call this H-sequence the Panov H-sequence of $E$.

Using the notation of section 2, for $j=1,\dots,s$, set :

[TABLE]

In [7], Panov proved that for $j=1,\dots,s$, $\mathfrak{n}_{j}$ and $\mathfrak{m}_{j}$ are Lie subalgebras. Consider the localization $S(\mathfrak{m}_{j-1})_{X_{\theta_{j}}}$ of the algebra $S(\mathfrak{m}_{j-1})$ with respect to the multiplicative subset generated by $X_{\theta_{j}}$. He defined an embedding of Poisson algebras $\Psi_{j-1}:S(\mathfrak{m}_{j})\rightarrow S(\mathfrak{m}_{j-1})_{X_{\theta_{j}}}$. Moreover, one observes directly from the definition of $\Psi_{j-1}$ that it is $\mathfrak{h}$-equivariant.

Extending the $\Psi_{j}$ with the appropriate localizations, we set

[TABLE]

Panov proved that the $f_{j}$’s are algebraically independent elements of $\Bbbk(\mathfrak{m}_{\Phi}^{*})^{\mathfrak{m}_{\Phi}}$. More precisely, we have $\Bbbk(\mathfrak{m}_{\Phi}^{*})^{\mathfrak{m}_{\Phi}}=\Bbbk(f_{1},\dots,f_{s})$, and hence $\chi(\mathfrak{m}_{\Phi})=s$. Furthermore, using the fact that the embeddings $\Psi_{j-1}$ are $\mathfrak{h}$-equivariant, the element $f_{j}$ is of weight $\theta_{j}$ for $j=1,\dots,s$.

Let ${\mathcal{I}}\subset\{1,\dots,s\}$ be such that $\{\theta_{i};i\in{\mathcal{I}}\}$ is a basis of $D(\mathbf{h})$. For $j\in\{1,\dots,s\}\setminus{\mathcal{I}}$, we have

[TABLE]

where $\lambda_{i}\in\mathbb{Z}$ for $i\in{\mathcal{I}}$ and $\lambda_{j}\in\mathbb{Z}^{*}$.

Set

[TABLE]

By construction, the elements $g_{j}$ are of weight zero. Hence $g_{j}\in\Bbbk(\mathfrak{m}_{\Phi}^{*})^{\mathfrak{q}_{\Phi}}$. Since the elements $f_{1},\dots,f_{s}$ are algebraically independent, it follows that

[TABLE]

Theorem 4.1**.**

Let $\mathbf{h}=(\theta_{1},\dots,\theta_{s})$ be the Panov H-sequence of $E$. There exists a non-empty open subset $U$ of $(\Bbbk^{*})^{s}$ such that $\dim\mathfrak{q}_{\Phi}^{f_{\mathbf{a}}}=c(\mathbf{h})$ whenever $\mathbf{a}\in U$. Moreover, we have

[TABLE]

Proof.

Let $U$ be a non-empty open subset of $(\Bbbk^{*})^{s}$ verifying part (iii) of Lemma 3.1. Set $r=d(\mathbf{h})$. Let ${\mathcal{I}}=\{i_{1},\dots,i_{r}\}\subset\{1,\dots,s\}$ be such that $(\theta_{i_{1}},\dots,\theta_{i_{r}})$ is a basis of $D(\mathbf{h})$ and complete to a basis $\mathcal{B}^{\prime}=(\beta_{1},\dots,\beta_{\ell})$ of $\mathfrak{h}^{*}$ such that $\beta_{k}=\theta_{i_{k}}$ for $k=1,\dots,r$. Denote by $\mathcal{B}=(h_{1},\dots,h_{\ell})$ the basis of $\mathfrak{h}$ dual to $\mathcal{B}^{\prime}$.

Let $m=\dim\mathfrak{m}_{\Phi}$ and $\mathcal{C}$ be a basis of $\mathfrak{m}_{\Phi}$. Then the matrix of $\Phi_{f_{\mathbf{a}}}$ in the basis $\mathcal{B}^{\prime}\cup\mathcal{C}$ is

[TABLE]

where $A$ is an element of rank $d(\mathbf{h})$ in the set of $\ell\times m$ matrices $\mathcal{M}_{\ell,m}(\Bbbk)$, and $B\in\mathcal{M}_{m,m}(\Bbbk)$ the set of $m\times m$ matrices. Set

[TABLE]

Then by (2), we have

[TABLE]

It follows that

[TABLE]

and since $\operatorname{rk}(M)\leqslant\operatorname{rk}(A)+\operatorname{rk}(M^{\prime})$, we deduce that

[TABLE]

Hence,

[TABLE]

It follows by Theorem 3.2 that $\dim\mathfrak{q}_{\Phi}^{f_{\mathbf{a}}}=c(\mathbf{h})$ and we have equalities in (3) and (4), (5). Consequently $\operatorname{rk}(M^{\prime})=\dim\mathfrak{m}_{\Phi}-\ell(\mathbf{h})+d(\mathbf{h})$ and $\chi(\mathfrak{q}_{\Phi},\mathfrak{m}_{\Phi})=\ell(\mathbf{h})-d(\mathbf{h}).$ ∎

Theorem 4.2**.**

Let $\mathbf{h}=(\theta_{1},\dots,\theta_{s})$ be the Panov H-sequence of $E$. Then we have $\chi(\mathfrak{q}_{\Phi})=\ell-\ell(\mathbf{h})+2d(\mathbf{h})$.

Proof.

Let $U$ be a non-empty open subset of $(\Bbbk^{*})^{s}$ verifying part (iii) of Lemma 3.1. Let $S$ be the subset of $\mathfrak{m}_{\Phi}^{*}$ consisting of elements of the form

[TABLE]

for $\boldsymbol{\lambda}=(\lambda_{1},\dots,\lambda_{s})\in\Bbbk^{s}$. Then $\Omega=\{f_{\mathbf{a}};\mathbf{a}\in U\}$ is an open subset of $S$.

Let $M$ be the algebraic adjoint group of $\mathfrak{m}_{\Phi}$. Consider the elements $z_{1},\dots,z_{s}$ of $\Bbbk[\mathfrak{m}_{\Phi}^{*}]^{M}$ constructed by Panov in [7] such that $\Bbbk(\mathfrak{m}_{\Phi}^{*})^{M}=\Bbbk(z_{1},\dots,z_{s})$, and

[TABLE]

where $P$ is some product of powers of $z_{1},\dots,z_{i-1}$ and $R$ is a polynomial in $X_{\alpha}$ for $\alpha\succ\theta_{i}$.

For $i=1,\dots,s$, denote by $U_{i}=\{f\in\mathfrak{q}_{\Phi}^{*};z_{i}(f)\not=0\}$ the standard open subset of $\mathfrak{q}_{\Phi}^{*}$ associated to $z_{i}$. By (6), $z_{1}=X_{\theta_{1}}$ so we clearly have $U_{1}\cap\Omega\not=\emptyset$. Next, for $i>0$, we have

[TABLE]

By the properties of $P$ and $R$ from the preceding paragraph, $P$ depends only on $z_{1},\dots,z_{i}$, and $R(f_{\boldsymbol{\lambda}})$ depends only on $\lambda_{1},\dots,\lambda_{i}$. By induction, we obtain

[TABLE]

Hence $\displaystyle\Omega^{\prime}=\Omega\cap\left(\bigcap_{i=1}^{s}U_{i}\right)$ is a non-empty open subset of $\Omega$.

Consider the map

[TABLE]

Assume that $f_{\boldsymbol{\lambda}}$ and $f_{\boldsymbol{\mu}}$ are two elements of $\Omega^{\prime}$ which are in the same $M$-orbit. Then we have $z_{i}(f_{\boldsymbol{\lambda}})=z_{i}(f_{\boldsymbol{\mu}})$ for $i=1,\dots,s$. In particular, we have $z_{1}(f_{\boldsymbol{\lambda}})=z_{1}(f_{\boldsymbol{\mu}})$ so $\lambda_{1}=\mu_{1}$.

Let us proceed by induction. Suppose that $i>0$ and that $\lambda_{j}=\mu_{j}$ for $1\leqslant j\leqslant i$. We have

[TABLE]

Since $f_{\boldsymbol{\lambda}},f_{\boldsymbol{\mu}}\in\Omega^{\prime}$, we deduce from the properties of $P$ and $R$ described above that $\lambda_{i+1}=\mu_{i+1}$. Hence $\boldsymbol{\lambda}=\boldsymbol{\mu}$.

It follows that for any $f_{\boldsymbol{\lambda}}\in\Omega^{\prime}$, we have

[TABLE]

By [7] and Theorem 3.2, we have $\dim\mathfrak{m}_{\Phi}^{f_{\boldsymbol{\lambda}}}=\ell(\mathbf{h})=s$. Since $\dim\mathfrak{m}_{\Phi}^{f_{\boldsymbol{\lambda}}}=\dim\mathrm{Stab}_{M}(f_{\boldsymbol{\lambda}})$ and $\dim\mathrm{Im}\Psi=\dim(M\times\Omega^{\prime})-\dim\Psi^{-1}(f_{\boldsymbol{\lambda}})$, we obtain that $\dim\mathrm{Im}\Psi=\dim M$. Therefore $M.\Omega^{\prime}$ contains an open subset $\mathcal{O}$ of $\mathfrak{m}_{\Phi}^{*}$.

Let $p$ be the projection $\mathfrak{q}_{\Phi}^{*}\rightarrow\mathfrak{m}_{\Phi}^{*}$ via restriction. Then $p$ is $M$-equivariant. Since the set of regular elements of $\mathfrak{q}_{\Phi}^{*}$ is an open subset, we deduce that there exist $\varphi\in p^{-1}(\mathcal{O})$ and $f_{\boldsymbol{\lambda}}\in\Omega^{\prime}$ such that $\varphi$ is regular and

[TABLE]

For all $X,Y\in\mathfrak{q}_{\Phi}$, we have $\varphi([X,Y])=f_{\boldsymbol{\lambda}}([X,Y])$, and therefore

[TABLE]

Hence by Theorem 4.1, we have

[TABLE]

∎

Remark 4.3**.**

According to the previous theorem, $c(\mathbf{h})$ is minimal when $\mathbf{h}$ is the Panov H-sequence.

Let $\mathfrak{g}$ be of type $A_{6}$. Set $\Phi=\{\alpha\in\Delta^{+};\alpha\geqslant\alpha_{2,5}\}$, and $E=\Delta^{+}\setminus\Phi$. Then the Panov H-sequence of $E$ is $\mathbf{h}=\{\alpha_{1,4},\alpha_{2,3},\alpha_{3,5},\alpha_{4,6},\alpha_{5,6},\alpha_{5,5}\}$ and we have $\ell(\mathbf{h})=d(\mathbf{h})=6$.

We have another H-sequence $\mathbf{h}^{\prime}=(\alpha_{3,5},\alpha_{4,6},\alpha_{6,6},\alpha_{1,4},\alpha_{1,3},\alpha_{1,2},\alpha_{2,3},$ $\alpha_{2,2})$ such that $\ell(\mathbf{h}^{\prime})=8$ and $d(\mathbf{h}^{\prime})=6$. Observe that $c(\mathbf{h})=0$ and $c(\mathbf{h}^{\prime})=2$.

5. Stability

Let $\mathfrak{a}$ be an algebraic Lie algebra and let $A$ be its adjoint algebraic group. Recall that $g\in\mathfrak{a}^{*}$ is stable if there exists an open subset $U$ of $\mathfrak{a}^{*}$ containing $g$ such that $\mathfrak{a}^{g}$ and $\mathfrak{a}^{h}$ are $A$-conjugate for all $h\in U$.

The following result is proved in [10] :

Proposition 5.1**.**

Let $\mathfrak{a}$ be an algebraic Lie algebra and $f\in\mathfrak{a}^{*}$.

1. (i)

If $f$ is stable, then it is a regular element of $\mathfrak{a}^{*}$. 2. (ii)

The linear form $f$ is stable if and only if $[\mathfrak{a},\mathfrak{a}^{f}]\cap\mathfrak{a}^{f}=\{0\}$.

In this section, we return to the general case, that is $\mathfrak{g}$ is not necessarily of type $A$.

Proposition 5.2**.**

Let $\Phi$ be a subset of $\Delta^{+}$ such that $\mathfrak{g}^{\Phi}$ is an ad-nilpotent ideal of $\mathfrak{g}$. Let $\mathbf{h}=(\theta_{1},\dots,\theta_{s})$ be an H-sequence of $\Delta^{+}\setminus\Phi$ consisting of linear independent elements. Then there exists $f\in\mathfrak{q}_{\Phi}^{*}$ which is stable and $\chi(\mathfrak{q}_{\Phi})=c(\mathbf{h})$.

Proof.

Our hypothesis implies that $\ell(\mathbf{h})=d(\mathbf{h})$. So

[TABLE]

Let $U$ be a non-empty open subset of $(\Bbbk^{*})^{s}$ verifying part (iii) of Lemma 3.1. By Theorem 3.2 and Lemma 3.1, if $\mathbf{a}\in U$, we have $\dim\mathfrak{q}_{\Phi}^{f_{\mathbf{a}}}=c(\mathbf{h})$. It follows from Lemma 3.1 that $\mathfrak{q}_{\Phi}^{f_{\mathbf{a}}}$ is a commutative Lie subalgebra of $\mathfrak{q}_{\Phi}$ consisting of semi-simple elements. Therefore, there exists a vector subspace $\mathfrak{r}$ of $\mathfrak{q}_{\Phi}$ such that $\mathfrak{q}_{\Phi}=\mathfrak{q}_{\Phi}^{f_{\mathbf{a}}}\oplus\mathfrak{r}$ and $[\mathfrak{q}_{\Phi}^{f_{\mathbf{a}}},\mathfrak{r}]\subset\mathfrak{r}$. So $[\mathfrak{q}_{\Phi},\mathfrak{q}_{\Phi}^{f_{\mathbf{a}}}]\subset\mathfrak{r}$ and the result follows by Theorem 5.1. ∎

We shall now show that $\mathfrak{q}_{\Phi}$ does not necessarily contain a stable linear form in general.

Let ${\mathcal{M}}_{6,6}(\Bbbk)$ be the set of $6\times 6$ matrices and let $\{E_{i,j},1\leqslant i,j\leqslant 6\}$ be its canonical basis. Let $\mathfrak{g}$ be the subset of ${\mathcal{M}}_{6,6}(\Bbbk)$ whose trace is equal to zero. We fix $\mathfrak{h}$ the set of diagonal matrices of $\mathfrak{g}$ and $\mathfrak{b}$ the set of upper triangular matrices of $\mathfrak{g}$.

Therefore, we can choose $X_{\alpha_{i}+\dots+\alpha_{j}}=E_{i,j+1}$, for $1\leqslant i\leqslant j\leqslant 5$. Set

[TABLE]

The Panov H-sequence of $\Delta^{+}\setminus\Phi$ is

[TABLE]

We have $\ell(\mathbf{h})=6$, $d(\mathbf{h})=5$ and $c(\mathbf{h})=1$.

By definition, $\mathfrak{i}=\mathfrak{g}^{\Phi}$ is an ad-nilpotent ideal of $\mathfrak{g}$.

Proposition 5.3**.**

The Lie algebra $\mathfrak{q}_{\Phi}^{*}$ does not possess any stable linear form.

Proof.

By Theorem 4.2 or Proposition 3.3, we have $\chi(\mathfrak{q}_{\Phi})=1$. Let $t\in\Bbbk^{*}$ and $\boldsymbol{\lambda}_{t}=(1,\dots,1,t)\in\Bbbk^{6}$. Set $f_{\boldsymbol{\lambda}_{t}}=\sum_{i=1}^{5}X_{\theta_{i}}+tX_{\theta_{6}}$ and

[TABLE]

A simple calculation gives $\mathfrak{q}_{\Phi}^{f_{\boldsymbol{\lambda}_{t}}}=\mathrm{Vect}(Z_{t}).$

Let $H\in\mathfrak{h}$ be such that $\alpha_{1}(H)=\alpha_{3}(H)=\alpha_{5}(H)=1$ and $\alpha_{2}(H)=\alpha_{4}(H)=0$. Then $[H,Z_{t}]=Z_{t}$, so $[\mathfrak{q}_{\Phi},\mathfrak{q}_{\Phi}^{f_{\boldsymbol{\lambda}_{t}}}]\cap\mathfrak{q}_{\Phi}^{f_{\boldsymbol{\lambda}_{t}}}\not=\{0\}$. By Theorem 5.1, $f_{\boldsymbol{\lambda}_{t}}$ is not stable.

Denote by $Q$ the algebraic adjoint group of $\mathfrak{q}_{\Phi}$. Then $Q$ can be identified with the quotient of the set of invertible upper triangular matrices by a closed normal subgroup. Let $s,t\in\Bbbk^{*}$. Assume that $\mathfrak{q}_{\Phi}^{f_{\boldsymbol{\lambda}_{t}}}$ and $\mathfrak{q}_{\Phi}^{f_{\boldsymbol{\lambda}_{s}}}$ are $Q$-conjugate, then there exist $\lambda\in\Bbbk^{*}$ and an invertible triangular matrix $P$ such that $PZ_{s}P^{-1}-\lambda Z_{t}\in\mathfrak{i}$. By the definition of $\mathfrak{i}$, for any element $L\in\mathfrak{i}$ and for any upper triangular matrix $R$, we have $LR\in\mathfrak{i}$. It follows that $PZ_{s}-\lambda Z_{t}P\in\mathfrak{i}$. By a direct computation, we obtain that $t=s$.

Recall that for $f,g\in\mathfrak{q}_{\Phi}^{*}$, if $f$ and $g$ are $Q$-conjugate, then $\mathfrak{q}_{\Phi}^{f}$ and $\mathfrak{q}_{\Phi}^{g}$ are also $Q$-conjugate. We define

[TABLE]

By the above consideration, we deduce that

[TABLE]

Since $\dim\mathrm{Im}\Psi=\dim(Q\times\Bbbk^{*})-\dim\Psi^{-1}(f_{\boldsymbol{\lambda}_{t}})$, we have $\dim\mathrm{Im}\Psi=\dim Q$. Therefore $Q.\Bbbk^{*}$ contains a non-empty open subset of $\mathfrak{q}_{\Phi}^{*}$ which does not contain any stable linear form.

Since the set of stable linear forms of $\mathfrak{q}_{\Phi}$ is an open subset of $\mathfrak{q}_{\Phi}^{*}$, the result follows immediately. ∎

6. Remarks on the exactness of the upper bounds

Assume that $\mathfrak{g}$ is of type $C_{7}$. Using the numbering of simple roots in [12], set

[TABLE]

We check by hand that the minimal length of an H-sequence associated to $\Delta^{+}\setminus\Phi$ is $8$. For example, the following H-sequence $\mathbf{h}=(\theta_{1},\dots,\theta_{s})$ is of length $8$ :

[TABLE]

and $d(\mathbf{h})=7$.

By Theorem 3.2, we have

[TABLE]

So $\chi(\mathfrak{q}_{\Phi})=1$ by Proposition 3.3. But by considering an arbitrary linear form, we found that $\chi(\mathfrak{m}_{\Phi})\leqslant 6$. Thus the upper bound for the index of $\mathfrak{m}_{\Phi}$ is not always exact when $\mathfrak{g}$ is not of type $A$.

We did some computations using Gap4 on arbitrary linear forms when $\mathfrak{g}$ is of rank less than or equal to $6$, and we have not found an example where the upper bound for $\chi(\mathfrak{q}_{\Phi})$ is not exact. This leads us to formulate the following conjecture :

Conjecture 6.1**.**

Let $\Phi\subset\Delta^{+}$ such that $\mathfrak{g}^{\Phi}$ is an ad-nilpotent ideal of $\mathfrak{g}$. There exists an H-sequence $\mathbf{h}$ of $\Delta^{+}\setminus\Phi$ such that

[TABLE]