---
abstract: 'Let $L$ be a reductive subgroup of a reductive Lie group $G$. Let $G/H$ be a homogeneous space of reductive type. We provide a necessary condition for the properness of the action of $L$ on $G/H$. As an application we give examples of spaces that do not admit standard compact Clifford-Klein forms.'
author:
- Maciej Bocheński and Marek Ogryzek
title: A restriction on proper actions on homogeneous spaces of a reductive type
---

Introduction
============

Let $L$ be a locally compact topological group acting continuously on a locally Hausdorff topological space $M$. This action is called [***proper***]{} if for every compact subset $C \subset M$ the set $$L(C):=\{  g\in L \ | \ g\cdot C \cap C \neq \emptyset \}$$ is compact. In this paper, our main concern is the following question posed by T. Kobayashi [@kob3]

How “large” subgroups of $G$ can act properly

on a homogeneous space $G/H$? **(Q1)**

We restrict our attention to the case where $M=G/H$ is a homogeneous space of reductive type and always assume that $G$ is a linear connected reductive real Lie group with the Lie algebra $\mathfrak{g}.$ Let $H\subset G$ be a closed subgroup of $G$ with finitely many connected components and $\mathfrak{h}$ be the Lie algebra of $H.$

The subgroup $H$ is reductive in $G$ if $\mathfrak{h}$ is reductive in $\mathfrak{g},$ that is, there exists a Cartan involution $\theta $ for which $\theta (\mathfrak{h}) = \mathfrak{h}.$ The space $G/H$ is called the homogeneous space of reductive type. \[def1\]

Note that if $\mathfrak{h}$ is reductive in $\mathfrak{g}$ then $\mathfrak{h}$ is a reductive Lie algebra.

It is natural to ask when does a closed subgroup of $G$ act properly on a space of reductive type $G/H.$ This problem was treated, inter alia, in [@ben], [@bt], [@kas], [@kob2], [@kob4], [@kob1], [@kul] and [@ok]. In [@kob2] one can find a very important criterion for a proper action of a subgroup $L$ reductive in $G.$ To state this criterion we need to introduce some additional notation. Let $\mathfrak{l}$ be the Lie algebra of $L.$ Take a Cartan involution $\theta$ of $\mathfrak{g}.$ We obtain the Cartan decomposition $$\mathfrak{g}=\mathfrak{k} + \mathfrak{p}.
 \label{eq1}$$ Choose a maximal abelian subspace $\mathfrak{a}$ in $\mathfrak{p}.$ The subspace $\mathfrak{a}$ is called the ***maximally split abelian subspace*** of $\mathfrak{p}$ and $\text{rank}_{\mathbb{R}}(\mathfrak{g}) := \text{dim} (\mathfrak{a})$ is called the ***real rank*** of $\mathfrak{g}.$ It follows from Definition \[def1\] that $\mathfrak{h}$ and $\mathfrak{l}$ admit Cartan decompositions $$\mathfrak{h}=\mathfrak{k}_{1} + \mathfrak{p}_{1} \ \text{and} \ \mathfrak{l}=\mathfrak{k}_{2} + \mathfrak{p}_{2},$$ given by Cartan involutions $\theta_{1}, \ \theta_{2}$ of $\mathfrak{g}$ such that $\theta_{1} (\mathfrak{h})= \mathfrak{h}$ and $\theta_{2} (\mathfrak{l})= \mathfrak{l}.$ Let $\mathfrak{a}_{1} \subset \mathfrak{p}_{1}$ and $\mathfrak{a}_{2} \subset \mathfrak{p}_{2}$ be maximally split abelian subspaces of $\mathfrak{p}_{1}$ and $\mathfrak{p}_{2},$ respectively. One can show that there exist $a,b \in G$ such that $\mathfrak{a}_{\mathfrak{h}} := \text{\rm Ad}_{a}\mathfrak{a}_{1} \subset \mathfrak{a}$ and $\mathfrak{a}_{\mathfrak{l}} := \text{\rm Ad}_{b}\mathfrak{a}_{2} \subset \mathfrak{a}.$ Denote by $W_{\mathfrak{g}}$ the Weyl group of $\mathfrak{g}.$ In this setting the following holds

The following three conditions are equivalent

1.  $L$ acts on $G/H$ properly.

2.  $H$ acts on $G/L$ properly.

3.  For any $w \in W_{\mathfrak{g}},$ $w\cdot \mathfrak{a}_{\mathfrak{l}} \cap \mathfrak{a}_{\mathfrak{h}} =\{ 0 \}.$

\[twkob\]

Note that the criterion 3. in Theorem \[twkob\] depends on how $L$ and $H$ are embedded in $G$ up to inner-automorphisms. Theorem \[twkob\] provides a partial answer to Q1.

If $L$ acts properly on $G/H$ then $$\text{\rm rank}_{\mathbb{R}}(\mathfrak{l}) + \text{\rm rank}_{\mathbb{R}}(\mathfrak{h}) \leq \text{\rm rank}_{\mathbb{R}} (\mathfrak{g}).$$ \[coko\]

Hence the real rank of $L$ is bounded by a constant which depends on $G/H,$ no matter how $H$ and $L$ are embedded in $G.$ In this paper we find a similar, stronger restriction for Lie groups $G,H,L$ by means of a certain tool which we call the a-hyperbolic rank (see Section 2, Definition \[dd2\] and Table \[tab1\]). In more detail we prove the following

If $L$ acts properly on $G/H$ then $$\mathop{\mathrm{rank}}\nolimits_{\text{\rm a-hyp}}(\mathfrak{l}) + \mathop{\mathrm{rank}}\nolimits_{\text{\rm a-hyp}}(\mathfrak{h}) \leq \mathop{\mathrm{rank}}\nolimits_{\text{\rm a-hyp}} (\mathfrak{g}).$$ \[twgl\]

Recall that a homogeneous space $G/H$ of reductive type admits a ***compact Clifford-Klein form*** if there exists a discrete subgroup $\Gamma \subset G$ such that $\Gamma$ acts properly on $G/H$ and $\Gamma \backslash G/H$ is compact. The space $G/H$ admits a ***standard compact Clifford-Klein form*** in the sense of Kassel-Kobayashi [@kako] if there exists a subgroup $L$ reductive in $G$ such that $L$ acts properly on $G/H$ and $L \backslash G/H$ is compact. In the latter case, for any discrete cocompact subgroup $\Gamma ' \subset L,$ the space $\Gamma ' \backslash G/H$ is a compact Clifford-Klein form. Therefore it follows from Borel’s theorem (see [@bor]) that any homogeneous space of reductive type admitting a standard compact Clifford-Klein form also admits a compact Clifford-Klein form. It is not known if the converse statement holds, but all known reductive homogeneous spaces $G/H$ admitting compact Clifford-Klein forms also admit standard compact Clifford-Klein forms.

As a corollary to Theorem \[twgl\], we get examples of the semisimple symmetric spaces without standard compact Clifford-Klein forms. In particular, we cannot find the first example in the existing literature:

The homogeneous spaces $G/H=SL(2k+1, \mathbb{R})/SO(k-1,k+2)$ and $G/H=SL(2k+1, \mathbb{R})/Sp(k-1,\mathbb{R})$ for $k \geq 5$ do not admit standard compact Clifford-Klein forms. \[co1\]

Let us mention the following results, related to the above corollary.

-   T. Kobayashi proved in [@kobadm] that $SL(2k,\mathbb{R})/SO(k,k)$ for $k\geq 1$ and $SL(n,\mathbb{R})/Sp(l,\mathbb{R})$ for $0<2l \leq n-2$ do not admit compact Clifford-Klein forms.

-   Y. Benoist proved in [@ben] that $SL(2k+1,\mathbb{R})/SO(k,k+1)$ for $k\geq 1$ does not admit compact Clifford-Klein forms.

-   Y. Morita proved recently in [@mor] that $SL(p+q,\mathbb{R})/SO(p,q)$ does not admit compact Clifford-Klein forms if $p$ and $q$ are both odd.

Note that these works are devoted to the problem of existence of compact Clifford-Klein forms on a given homogeneous space (not only standard compact Clifford-Klein forms).

The a-hyperbolic rank and antipodal hyperbolic orbits
=====================================================

Let $\Sigma_{\mathfrak{g}}$ be a system of restricted roots for $\mathfrak{g}$ with respect to $\mathfrak{a}.$ Choose a system of positive roots $\Sigma^{+}_{\mathfrak{g}}$ for $\Sigma_{\mathfrak{g}}.$ Then a fundamental domain of the action of $W_{\mathfrak{g}}$ on $\mathfrak{a}$ can be define as $$\mathfrak{a}^{+} := \{  X\in \mathfrak{a} \ |  \ \alpha (X) \geq 0 \ \text{\rm for any} \ \alpha \in \Sigma^{+}_{\mathfrak{g}} \}.$$ Note that $$sX+tY \in \mathfrak{a}^{+},$$ for any $s,t \geq 0$ and $X,Y\in \mathfrak{a}^{+}$. Therefore $\mathfrak{a}^{+}$ is a convex cone in the linear space $\mathfrak{a}.$ Let $w_{0} \in W_{\mathfrak{g}}$ be the longest element. One can show that $$-w_{0}: \mathfrak{a} \rightarrow \mathfrak{a}, \ \ X \mapsto -(w_{0} \cdot X)$$ is an involutive automorphism of $\mathfrak{a}$ preserving $\mathfrak{a}^{+}.$ Let $\mathfrak{b} \subset \mathfrak{a}$ be the subspace of all fixed points of $-w_{0}$ and put $$\mathfrak{b}^{+} := \mathfrak{b} \cap \mathfrak{a}^{+}.$$ Thus $\mathfrak{b}^{+}$ is a convex cone in $\mathfrak{a}.$ We also have $\mathfrak{b} = \text{Span} (\mathfrak{b}^{+}).$

The dimension of $\mathfrak{b}$ is called the a-hyperbolic rank of $\mathfrak{g}$ and is denoted by $$\mathop{\mathrm{rank}}\nolimits_{\text{a-hyp}} (\mathfrak{g}).$$ \[dd2\]

The a-hyperbolic ranks of the simple real Lie algebras can be deduce from Table \[tab1\].

A method of calculation of the a-hyperbolic rank of a simple Lie algebra can be found in [@bt]. The a-hyperbolic rank of a semisimple Lie algebra equals the sum of a-hyperbolic ranks of all its simple parts. For a reductive Lie algebra $\mathfrak{g}$ we put $$\mathop{\mathrm{rank}}\nolimits_{\text{a-hyp}} (\mathfrak{g}) := \mathop{\mathrm{rank}}\nolimits_{\text{a-hyp}}([\mathfrak{g},\mathfrak{g}]).$$

There is a close relation between $\mathfrak{b}^{+}$ and the set of antipodal hyperbolic orbits in $\mathfrak{g}.$ We say that an element $X \in \mathfrak{g}$ is [***hyperbolic***]{}, if $X$ is semisimple (that is, $\mathrm{ad}_{X}$ is diagonalizable) and all eigenvalues of $\mathrm{ad}_{X}$ are real.

An adjoint orbit $O_{X}:=\mathop{\mathrm{Ad}}\nolimits (G)(X)$ is said to be hyperbolic if $X$ (and therefore every element of $O_{X}$) is hyperbolic. An adjoint orbit $O_{Y}$ is antipodal if $-Y\in O_{Y}$ (and therefore for every $Z\in O_{Y},$ $-Z\in O_{Y}$).

There is a bijective correspondence between antipodal hyperbolic orbits $O_{X}$ in $\mathfrak{g}$ and elements $Y \in \mathfrak{b}^{+}.$ This correspondence is given by $$\mathfrak{b}^{+}\ni Y \mapsto O_{Y}.$$ Furthermore, for every hyperbolic orbit $O_{X}$ in $\mathfrak{g}$ the set $O_{X} \cap \mathfrak{a}$ is a single $W_{\mathfrak{g}}$ orbit in $\mathfrak{a}$. \[lma\]

The main result
===============

We need two basic facts from linear algebra.

Let $V_{1},V_{2}$ be vector subspaces of a real linear space $V$ of finite dimension. Then $$\text{\rm dim} (V_{1}+V_{2})= \text{\rm dim} (V_{1}) + \text{\rm dim} (V_{2}) - \text{\rm dim} (V_{1}\cap V_{2}).$$ \[lma1\]

Let $V_{1},...,V_{n}$ be a collection of vector subspaces of a real linear space $V$ of a finite dimension and let $A^{+} \subset V$ be a convex cone. Assume that $$A^{+} \subset \bigcup_{k=1}^{n} V_{k}.$$ Then there exists a number $k,$ such that $A^{+} \subset V_{k}.$ \[lma2\]

We also need the following, technical lemma. Choose a subalgebra $\mathfrak{h}$ reductive in $\mathfrak{g}$ which corresponds to a Lie group $H \subset G.$ Let $\mathfrak{b}_{[\mathfrak{h},\mathfrak{h}]}^{+} \subset \mathfrak{a}_{\mathfrak{h}}$ be the convex cone constructed according to the procedure described in the previous subsection (for $[\mathfrak{h},\mathfrak{h}]$).

Let $X\in \mathfrak{b}_{[\mathfrak{h},\mathfrak{h}]}^{+}.$ The orbit $O_{X}:=\mathop{\mathrm{Ad}}\nolimits (G)(X)$ is an antipodal hyperbolic orbit in $\mathfrak{g}.$ \[lma3\]

By Lemma \[lma\] the vector $X$ defines an antipodal hyperbolic orbit in $\mathfrak{h}.$ Therefore we can find $h \in H \subset G$ such that $\text{\rm Ad}_{h}(X) = - X$. Since a maximally split abelian subspace $\mathfrak{a} \subset \mathfrak{g}$ consists of vectors for which $\mathrm{ad}$ is diagonalizable with real values and $$X \in \mathfrak{b}_{[\mathfrak{h},\mathfrak{h}]}^{+} \subset \mathfrak{a}_{\mathfrak{h}} \subset \mathfrak{a},$$ thus the vector $X$ is hyperbolic in $\mathfrak{g}.$ It follows that $\mathop{\mathrm{Ad}}\nolimits (G)(X)$ is a hyperbolic orbit in $\mathfrak{g}$ and $-X \in \mathop{\mathrm{Ad}}\nolimits (G)(X).$

Now we are ready to give a proof of Theorem \[twgl\].

Assume that $\mathop{\mathrm{rank}}\nolimits_{\text{a-hyp}}(\mathfrak{l}) + \mathop{\mathrm{rank}}\nolimits_{\text{a-hyp}}(\mathfrak{h}) > \mathop{\mathrm{rank}}\nolimits_{\text{a-hyp}} (\mathfrak{g})$ and let $\mathfrak{b}_{[\mathfrak{h},\mathfrak{h}]}^{+},$ $\mathfrak{b}_{[\mathfrak{l},\mathfrak{l}]}^{+},$ $\mathfrak{b}^{+}$ be appropriate convex cones. If $X \in \mathfrak{b}_{[\mathfrak{h},\mathfrak{h}]}^{+}$ then $O_{X}^{H} := \mathop{\mathrm{Ad}}\nolimits (G)(X)$ is an antipodal hyperbolic orbit in $\mathfrak{h}.$ By Lemma \[lma3\] the orbit $O_{X}^{G} := \mathop{\mathrm{Ad}}\nolimits (G)(X)$ is an antipodal hyperbolic orbit in $\mathfrak{g}.$ By Lemma \[lma\] there exists $Y \in \mathfrak{b}^{+}$ such that $$O_{X}^{G} = O_{Y}^{G} = \mathop{\mathrm{Ad}}\nolimits (G)(Y).$$ Since $\mathfrak{b}_{[\mathfrak{h},\mathfrak{h}]}^{+} \subset \mathfrak{a}_{\mathfrak{h}} \subset \mathfrak{a}$ and $\mathfrak{b}^{+} \subset \mathfrak{a}$ thus (according to Lemma \[lma\]) we get $X=w_{1} \cdot Y$ for a certain $w_{1} \in W_{\mathfrak{g}}.$ Therefore $$\mathfrak{b}_{[\mathfrak{h},\mathfrak{h}]}^{+} \subset W_{\mathfrak{g}} \cdot \mathfrak{b}^{+} = \bigcup_{w\in W_{\mathfrak{g}}} w \cdot \mathfrak{b}^{+} \subset \bigcup_{w\in W_{\mathfrak{g}}} w \cdot \mathfrak{b}.$$ Analogously $$\mathfrak{b}_{[\mathfrak{l},\mathfrak{l}]}^{+} \subset \bigcup_{w\in W_{\mathfrak{g}}} w \cdot \mathfrak{b}^{+} \subset \bigcup_{w\in W_{\mathfrak{g}}} w \cdot \mathfrak{b}.$$ By Lemma \[lma2\] there exist $w_{\mathfrak{h}},w_{\mathfrak{l}} \in W_{\mathfrak{g}}$ such that $$\mathfrak{b}_{[\mathfrak{h},\mathfrak{h}]}^{+} \subset w_{\mathfrak{h}}^{-1} \cdot \mathfrak{b} \ \ \text{\rm and} \ \ \mathfrak{b}_{[\mathfrak{l},\mathfrak{l}]}^{+} \subset w_{\mathfrak{l}}^{-1} \cdot \mathfrak{b},$$ because $W_{\mathfrak{g}}$ acts on $\mathfrak{a}$ by linear transformations. Therefore $$\mathfrak{b}_{[\mathfrak{h},\mathfrak{h}]} \subset w_{\mathfrak{h}}^{-1} \cdot \mathfrak{b} \ \ \text{\rm and} \ \ \mathfrak{b}_{[\mathfrak{l},\mathfrak{l}]} \subset w_{\mathfrak{l}}^{-1} \cdot \mathfrak{b}$$ where $\mathfrak{b}_{[\mathfrak{h},\mathfrak{h}]} := \text{Span}(\mathfrak{b}_{[\mathfrak{h},\mathfrak{h}]}^{+})$ and $\mathfrak{b}_{[\mathfrak{l},\mathfrak{l}]} := \text{Span} (\mathfrak{b}_{[\mathfrak{l},\mathfrak{l}]}^{+}).$ We obtain $$w_{\mathfrak{h}} \cdot \mathfrak{b}_{[\mathfrak{h},\mathfrak{h}]} \subset \mathfrak{b} , \ w_{\mathfrak{l}} \cdot \mathfrak{b}_{[\mathfrak{l},\mathfrak{l}]} \subset \mathfrak{b}.$$ By the assumption and Lemma \[lma1\] $$\text{\rm dim}(w_{\mathfrak{h}} \cdot \mathfrak{b}_{[\mathfrak{h},\mathfrak{h}]} \cap w_{\mathfrak{l}} \cdot \mathfrak{b}_{[\mathfrak{l},\mathfrak{l}]}) >0.$$ Choose $0 \neq Y \in w_{\mathfrak{h}} \cdot \mathfrak{b}_{[\mathfrak{h},\mathfrak{h}]} \cap w_{\mathfrak{l}} \cdot \mathfrak{b}_{[\mathfrak{l},\mathfrak{l}]}.$ Then $$w_{\mathfrak{l}} \cdot X_{\mathfrak{l}}=Y= w_{\mathfrak{h}} \cdot X_{\mathfrak{h}} \ \text{\rm for some} \ X_{\mathfrak{h}} \in \mathfrak{b}_{[\mathfrak{h},\mathfrak{h}]}\backslash \{ 0 \}  \ \text{\rm and} \ X_{\mathfrak{l}} \in \mathfrak{b}_{[\mathfrak{l},\mathfrak{l}]}\backslash \{  0 \}.$$ Take $w_{2} := w_{\mathfrak{h}}^{-1}w_{\mathfrak{l}} \in W_{\mathfrak{g}},$ we have $X_{\mathfrak{h}}= w_{2} \cdot X_{\mathfrak{l}}$ and $X_{\mathfrak{h}} \in \mathfrak{a}_{\mathfrak{h}}, \ X_{\mathfrak{l}} \in \mathfrak{a}_{\mathfrak{l}}.$ Thus $0 \neq X_{\mathfrak{h}} \in w_{2} \cdot \mathfrak{a}_{\mathfrak{l}} \cap \mathfrak{a}_{\mathfrak{h}}.$ The assertion follows from Theorem \[twkob\].

We can proceed to a proof of Corollary \[co1\]. For a reductive Lie group $D$ with a Lie algebra $\mathfrak{d}$ with a Cartan decomposition $$\mathfrak{d} = \mathfrak{k}_{\mathfrak{d}} + \mathfrak{p}_{\mathfrak{d}}$$ we define $d(G) := \text{dim} (\mathfrak{p}_{\mathfrak{d}}).$ We will need the following properties

Let $L$ be a subgroup reductive in $G$ acting properly on $G/H.$ The space $L \backslash G /H$ is compact if and only if $$d(L)+d(H)=d(G).$$ \[twkk\]

If $J \subset G$ is a semisimple subgroup then it is reductive in $G.$ \[twy\]

Let $L \subset G$ be a semisimple Lie group acting properly on $G/H=SL(2k+1, \mathbb{R})/SO(k-1,k+2)$ or $G/H=SL(2k+1, \mathbb{R})/Sp(k-1,\mathbb{R}).$ Then $$\text{\rm rank}_{\mathbb{R}}(\mathfrak{l}) \leq 2.$$ \[p1\]

Because $\mathop{\mathrm{rank}}\nolimits_{\text{a-hyp}}(\mathfrak{g}) = 1+ \mathop{\mathrm{rank}}\nolimits_{\text{a-hyp}}(\mathfrak{h})$ thus it follows from Table \[tab1\] and Theorem \[twgl\] that if $L$ is simple then $\text{rank}_{\mathbb{R}}(\mathfrak{l}) \leq 2.$ On the other hand if $L$ is semisimple then each (non-compact) simple part of $\mathfrak{l}$ adds at least $1$ to a-hyperbolic rank of $\mathfrak{l}.$ Thus we also have $\text{rank}_{\mathbb{R}}(\mathfrak{l}) \leq 2.$

*Proof of Corollary \[co1\].* Assume now that $L$ is reductive in $G.$ Since the Lie algebra $\mathfrak{l}$ is reductive therefore $$\mathfrak{l} = \mathfrak{c}_{\mathfrak{l}} + [\mathfrak{l},\mathfrak{l}],$$ where $\mathfrak{c}_{\mathfrak{l}}$ denotes the center of $\mathfrak{l}.$ It follows from Corollary \[coko\] that $$\text{\rm rank}_{\mathbb{R}}(\mathfrak{l}) \leq k+1,
\label{eq2}$$ and by Proposition \[p1\] we have $\text{rank}_{\mathbb{R}}([\mathfrak{l},\mathfrak{l}]) \leq 2.$ Note that $$d(G)-d(H)\geq k^{2} +2k +2.$$ We will show that if $L$ acts properly on $G/H$ and $k\geq 5$ then $$d(L) < k^{2} + 2k +2.
\label{eq4}$$ Let $[\mathfrak{l},\mathfrak{l}] = \mathfrak{k}_{0} + \mathfrak{p}_{0}$ be a Cartan decomposition. From (\[eq2\]) $$d(L) \leq \text{\rm dim} (\mathfrak{c}_{\mathfrak{l}}) + \text{\rm dim} (\mathfrak{p}_{0}) \leq k+1 + \text{\rm dim} (\mathfrak{p}_{0}).
\label{eq7}$$ Also, if $\text{rank}_{\mathbb{R}}([\mathfrak{l},\mathfrak{l}]) =2$ then it follows from Table \[tab1\] that (the only) non-compact simple part of $[\mathfrak{l},\mathfrak{l}]$ is isomorphic to $\mathfrak{sl}(3,\mathbb{R}),$ $\mathfrak{su}^{\ast}(6),$ $\mathfrak{e}_{6}^{\text{IV}}$ or $\mathfrak{sl}(3,\mathbb{C})$ (treated as a simple real Lie algebra). In such case $$\text{\rm dim} (\mathfrak{p}_{0}) < 27.
\label{eq5}$$ Therefore assume that $\text{rank}_{\mathbb{R}} ([\mathfrak{l},\mathfrak{l}])=1$ and let $\mathfrak{s} \subset [\mathfrak{l},\mathfrak{l}]$ be (the only) simple part of a non-compact type. We have $$\text{\rm rank}_{\mathbb{R}} (\mathfrak{s}) =1.
\label{eq3}$$ It follows from Theorem \[twy\] that $\mathfrak{s}$ is reductive in $\mathfrak{g}.$ Therefore $\mathfrak{s}$ admits a Cartan decomposition $$\mathfrak{s} = \mathfrak{k}_{\mathfrak{s}} + \mathfrak{p}_{\mathfrak{s}}$$ compatible with $\mathfrak{g}= \mathfrak{k} + \mathfrak{p},$ that is $\mathfrak{k}_{\mathfrak{s}} \subset \mathfrak{k}.$ We also have $\text{dim}(\mathfrak{p}_{s})=\text{dim}(\mathfrak{p}_{0}).$ Since $\mathfrak{k} = \mathfrak{so}(2k+1)$ we obtain $$\text{\rm rank} (\mathfrak{k}_{s}) \leq \text{\rm rank} (\mathfrak{k}) = k.$$ Using the above condition together with (\[eq3\]) we can check (by a case-by-case study of simple Lie algebras) that $$\text{\rm dim} (\mathfrak{p}_{\mathfrak{s}}) < 4k.
\label{eq6}$$ Now (\[eq7\]), (\[eq5\]) and (\[eq6\]) imply that $$d(L) < 5k+1$$ for $k \geq 6,$ and $d(L)<33$ for $k=5.$ Thus we have showed (\[eq4\]). The assertion follows from Theorem \[twkk\].

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Maciej Bocheński

Department of Mathematics and Computer Science,

University of Warmia and Mazury,

Słoneczna 54, 10-710, Olsztyn, Poland.

email: mabo@matman.uwm.edu.pl

Marek Ogryzek

Department of Geodesy and Land Management,

University of Warmia and Mazury,

Prawocheńskiego 15, 10-720, Olsztyn, Poland.

email: marek.ogryzek@uwm.edu.pl
