# Codimension two polar homogeneous foliations on symmetric spaces of
noncompact type

José Carlos Díaz-Ramos  and  Juan Manuel Lorenzo-Naveiro  Department of
Mathematics, CITMAGA, University of Santiago de Compostela, Spain
<EMAIL_ADDRESS><EMAIL_ADDRESS>
###### Abstract.

We classify homogeneous polar foliations of codimension two on irreducible
symmetric spaces of noncompact type up to orbit equivalence. Any such
foliation is either hyperpolar or the canonical extension of a polar
homogeneous foliation on a rank one symmetric space.

###### Key words and phrases:

Polar action, foliation, symmetric space

###### 2010 Mathematics Subject Classification:

53C35, 53C12, 57S20, 57S25

Supported by projects PID2019-105138GB-C21/AEI/10.13039/501100011033 (Spain)
and ED431C 2019/10, ED431F 2020/04 (Xunta de Galicia, Spain). The second
author acknowledges support of the FPI program (Spain).

##  1\. Introduction

A proper isometric action of a Lie group $G$ on a Riemannian manifold $M$ is
said to be _polar_ if there exists a connected and complete submanifold
$\Sigma$ that intersects every orbit orthogonally. The action is _hyperpolar_
if it admits a flat section (with the induced metric). Many results in Algebra
and Geometry can be thought of in terms of polar actions. For instance, the
system of polar coordinates in the Euclidean plane $\mathbb{R}^{2}$ is
connected to the polarity of the standard representation of the orthogonal
group $\mathsf{O}(2)$ in the plane. In a similar fashion, the spectral theorem
for hermitian matrices can be restated as follows: the action of the unitary
group $\mathsf{U}(n)$ on the vector space of $n\times n$ hermitian matrices by
conjugation is hyperpolar; a section is given by the subspace of diagonal
matrices. Based on this last example, we usually refer to the points of a
section as canonical forms of the elements of the ambient space (see [21] for
more details). A final example with a more Lie-theoretic flavor can be seen in
the adjoint action of a compact Lie group, whose sections are the maximal
tori.

Broadly speaking, we should not expect a generic Riemannian manifold to admit
a nontrivial polar action. Indeed, sections are known to be totally geodesic
submanifolds of the ambient space, so the existence of polar actions implies
the existence of (sufficiently large) families of globally defined Killing
fields and totally geodesic submanifolds orthogonal to each other; both
objects are rare in a space with no restrictions [20, Theorem A]. Thus, it
should not come as a surprise that polar actions can only be possible in
Riemannian spaces with a certain degree of symmetry. Some structural results
concerning spaces that do admit polar actions can be seen for example in [11,
Theorem A], and [17, Theorem 1], and its subsequent generalization in [22,
Theorem 1.1]. Consequently, the most natural classes of Riemannian manifolds
where we can develop a fruitful study of these actions are those of
homogeneous and symmetric spaces.

Our main interest in this paper concerns the classification of polar actions
(up to orbit equivalence) on manifolds with symmetries. Dadok [7] proved that
any polar representation on a Euclidean space is orbit equivalent to the
isotropy representation of a symmetric space, thus solving the problem in the
round sphere $\mathsf{S}^{n}$. The classification in Euclidean spaces follows
easily from Dadok’s result. The corresponding classification of polar actions
in the real hyperbolic space $\mathbb{R}\mathsf{H}^{n}$ was given by Wu in
[25].

As for symmetric spaces, while the problem is nearing its full conclusion in
the compact setting (see [16] and the references therein), little is known
about polar actions on noncompact symmetric spaces, our main interest for this
article. Nevertheless, there are already complete classifications in real and
complex hyperbolic spaces [8]. For the special case of cohomogeneity one
actions, we have the works of Berndt and Tamaru (see [4] and the references in
it) as well as a more recent structural result given by the first author,
Domínguez-Vázquez and Otero [9].

We say that the orbits of a proper action form a _homogeneous foliation_ if
they all have the same orbit type. Cohomogeneity one foliations have been
classified up to orbit equivalence in irreducible noncompact symmetric spaces
by Berndt and Tamaru in [3]. This result has been extended by Solonenko [23]
for the reducible case. In addition, Berndt, Tamaru and the first author give
in [2] a procedure to construct all possible hyperpolar homogeneous foliations
on any symmetric space of noncompact type; it is also shown that there are
polar foliations on noncompact irreducible symmetric spaces of higher rank
that are not hyperpolar. The results presented in these two papers make
extensive use of tools from the theory of real semisimple Lie algebras,
especially the root space and Iwasawa decompositions for these objects. At any
rate, we are yet to obtain general results concerning polar homogeneous
foliations that are not hyperpolar.

The aim of this paper is to start the study of polar nonhyperpolar homogeneous
foliations on symmetric spaces of noncompact type. We determine all of these
that have the hyperbolic plane $\mathbb{R}\mathsf{H}^{2}$ as a section. As a
consequence, combining this work with [2], we obtain a list of all polar
homogeneous foliations of codimension two in any such space. In order to state
the main result of this paper we need to introduce some concepts. See Section
2 for further details.

A Riemannian symmetric space of noncompact type can be written as a quotient
$M=G/K$, where $G$ is a semisimple Lie group and $K$ is the isotropy group at
a point $o\in M$. The Lie algebra $\mathfrak{g}$ of $G$ has a Cartan
decomposition $\mathfrak{g}=\mathfrak{k}\oplus\mathfrak{p}$, where
$\mathfrak{p}\cong T_{o}M$ is the orthogonal complement of $\mathfrak{k}$ in
$\mathfrak{g}$ with respect to the Killing form. We normalize the metric on
$M$ so that its restriction to $\mathfrak{p}$ coincides with the Killing form
of $\mathfrak{g}$ restricted to $\mathfrak{p}$. We choose a maximal abelian
subspace $\mathfrak{a}$ of $\mathfrak{p}$. This determines a root space
decomposition
$\mathfrak{g}=\mathfrak{g}_{0}\oplus\bigl{(}\bigoplus_{\lambda\in\Sigma}\mathfrak{g}_{\lambda}\bigr{)}$,
where $\Sigma$ is the set of roots with respect to $\mathfrak{a}$. We choose a
positivity criterion on $\Sigma$ and denote by $\Sigma^{+}$ the set of
positive roots. Let $\Lambda\subseteq\Sigma^{+}$ be the corresponding set of
simple roots. We define
$\mathfrak{n}=\bigoplus_{\lambda\in\Sigma^{+}}\mathfrak{g}_{\lambda}$. Then we
have an Iwasawa decomposition
$\mathfrak{g}=\mathfrak{k}\oplus\mathfrak{a}\oplus\mathfrak{n}$.

Let $\Phi\subseteq\Lambda$ be a subset of simple roots. Then $\Phi$ determines
a so-called parabolic subgroup $Q_{\Phi}$ with Langlands decomposition
$Q_{\Phi}=M_{\Phi}A_{\Phi}N_{\Phi}$, where $M_{\Phi}$ is reductive, $A_{\Phi}$
is abelian, and $N_{\Phi}$ is nilpotent. The totally geodesic submanifold
$B_{\Phi}=M_{\Phi}\cdot o$ is a symmetric space of noncompact type of rank
$\lvert\Phi\rvert$ that is called the boundary component associated with
$\Phi$. If $H_{\Phi}$ is a subgroup of the isometry group of $B_{\Phi}$, then
$H_{\Phi}A_{\Phi}N_{\Phi}$ induces an isometric action on $M$. We call this
action the canonical extension of the isometric action on the boundary
component $B_{\Phi}$ to the symmetric space $M$. See [4] for further details.

The main result of this paper is the following:

######  Theorem A.

Every codimension two polar nonhyperpolar homogeneous foliation on $M$ is
orbit equivalent to the canonical extension of a codimension two polar
nonhyperpolar homogeneous foliation on a boundary component of rank one in
$M$.

More explicitly, we prove:

######  Theorem B.

Let $M=G/K$ be a connected Riemannian symmetric space of noncompact type.
Then, a codimension two homogeneous polar nonhyperpolar foliation on $M$ is
orbit equivalent to the orbit foliation of the closed connected Lie group
whose Lie algebra is given by one of the following possibilities:

  1. (i)

$(\ker\alpha)\oplus(\mathfrak{n}\ominus\ell_{\alpha})$, where
$\alpha\in\Lambda$ is a simple root, and $\ell_{\alpha}$ is a line in
$\mathfrak{g}_{\alpha}$, or

  2. (ii)

$\mathfrak{a}\oplus(\mathfrak{n}\ominus\mathfrak{v}_{\alpha})$, where
$\alpha\in\Lambda$ is a simple root, and $\mathfrak{v}_{\alpha}$ is a
2-dimensional abelian subspace of $\mathfrak{g}_{\alpha}$.

We will show in Section 3 that different choices of $\ell_{\alpha}$ or
$\mathfrak{v}_{\alpha}$ above give rise to congruent foliations. We also
determine the mean curvature of their leaves. A direct consequence of this
computation is that

######  Corollary 1.1.

If $\mathcal{F}$ is a codimension two polar nonhyperpolar homogeneous
foliation on $M$, then $\mathcal{F}$ is harmonic if and only if $\mathcal{F}$
is orbit equivalent to the canonical extension of the trivial action on a
boundary component homothetic to the hyperbolic plane
$\mathbb{R}\mathsf{H}^{2}$.

We say that a foliation $\mathcal{F}$ is harmonic if all of its leaves are
minimal submanifolds of $M$. Equivalently, $\mathcal{F}$ is harmonic when the
canonical projection from $M$ to the space of leaves of $\mathcal{F}$ is a
harmonic map.

As a result of combining Theorem B with [2], Corollary 1.2 states the complete
classification of homogeneous polar foliations of codimension two. Recall that
two roots $\alpha$, $\beta\in\Lambda$ are said to be disconnected if
$\alpha+\beta$ is not a root (equivalently, if they are orthogonal).

######  Corollary 1.2.

A codimension two homogeneous polar foliation on $M$ is orbit equivalent to
the orbit foliation of a closed connected Lie group whose Lie algebra is:

  1. (a)

$(\mathfrak{a}\ominus\mathfrak{v})\oplus\mathfrak{n}$, where $\mathfrak{v}$ is
a $2$-dimensional subspace of $\mathfrak{a}$, or

  2. (b)

$(\mathfrak{a}\ominus\ell)\oplus(\mathfrak{n}\ominus\ell_{\alpha})$, where
$\alpha\in\Lambda$ is a simple root, $\ell_{\alpha}$ is a line in
$\mathfrak{g}_{\alpha}$, and $\ell$ is a line in $\ker\alpha$, or

  3. (c)

$\mathfrak{a}\oplus\bigl{(}\mathfrak{n}\ominus(\ell_{\alpha}\oplus\ell_{\beta})\bigr{)}$,
where $\alpha$, $\beta\in\Lambda$ are orthogonal simple roots, and
$\ell_{\lambda}$ is a line in $\mathfrak{g}_{\lambda}$,
$\lambda\in\\{\alpha,\beta\\}$, or

  4. (d)

$(\ker\alpha)\oplus(\mathfrak{n}\ominus\ell_{\alpha})$, where
$\alpha\in\Lambda$ is a simple root, and $\ell_{\alpha}$ is a line in
$\mathfrak{g}_{\alpha}$, or

  5. (e)

$\mathfrak{a}\oplus(\mathfrak{n}\ominus\mathfrak{v}_{\alpha})$, where
$\alpha\in\Lambda$ is a simple root, and $\mathfrak{v}_{\alpha}$ is a
2-dimensional abelian subspace of $\mathfrak{g}_{\alpha}$.

Examples (a) to (c) of Corollary 1.2 are hyperpolar.

######  Remark 1.3.

Recall that there is an implicit assumption on the metric imposed on $M$ so
that it coincides with the Killing form at $o$. In the case that $M$ is
irreducible, Schur’s lemma implies that all symmetric metrics on $M$ differ by
a scalar, and thus the classification remains true independently of the chosen
metric. If $M$ is reducible, the symmetric metrics are given by rescaling each
De Rham factor of $M$ by (possibly different) positive constants. As a
consequence, an isometric action that is polar and of cohomogeneity greater
than one may not admit a section upon alterations of the metric.

The trivial action is always polar and the whole space is a section of this
action. On the other hand, cohomogeneity one polar actions are automatically
hyperpolar. An easy consequence of Theorem B is that the action corresponding
to case (i) exists and is nontrivial unless $\Sigma^{+}=\\{\alpha\\}$ and
$\dim\mathfrak{g}_{\alpha}=1$. Thus,

######  Corollary 1.4.

If $M$ is an irreducible symmetric space of noncompact type where all polar
actions are hyperpolar, then $M$ is the real hyperbolic space
$\mathbb{R}\mathsf{H}^{2}$.

This contrasts sharply with the situation in the compact setting: polar
actions on irreducible symmetric spaces of compact type and higher rank are
always hyperpolar. This follows from a series of papers by Kollross that
concluded in the paper [16] by Kollross and Lytchak.

We now describe the structure of this paper. In Section 2 we review the basic
theory of real semisimple Lie algebras and symmetric spaces that is needed for
our purposes. In Section 3, we present the list of codimension two polar
nonhyperpolar homogeneous foliations that appear in Theorem B. We also
determine the curvature of their sections, the extrinsic geometry of their
orbits, and prove Theorem A. Finally, Section 4 contains the proof of Theorem
B.

##  2\. Preliminaries

In this section we introduce the concepts, notations, and preliminary results
that are used throughout this paper. We follow [13] for the theory on
symmetric spaces, and [15] for semisimple Lie algebras. Since symmetric spaces
of noncompact type are Hadamard manifolds, another interesting source of
information is [10].

###  2.1. Semisimple Lie algebras and symmetric spaces

A connected Riemannian symmetric space of noncompact type can be represented
as a quotient $M=G/K$, where $(G,K)$ is a symmetric pair. The group $G$ acts
almost effectively on $M$, and $K$ is taken to be the isotropy group of $G$ at
a point $o\in M$ that we fix from now on. Thus, the group $G$ is semisimple
and $K$ is a maximal compact subgroup of $G$. The Lie algebra of $G$ is
denoted by $\mathfrak{g}$. The Killing form $\mathcal{B}$ of $\mathfrak{g}$ is
nondegenerate because $\mathfrak{g}$ is a real semisimple Lie algebra;
$\mathcal{B}$ is known to be negative definite on $\mathfrak{k}$, the Lie
algebra of $K$. Let $\mathfrak{p}$ denote the orthogonal complement of
$\mathfrak{k}$ in $\mathfrak{g}$ with respect to the Killing form. Then,
$\mathfrak{g}=\mathfrak{k}\oplus\mathfrak{p}$ is a Cartan decomposition of
$\mathfrak{g}$, and $\mathfrak{p}$ can be identified with the tangent space
$T_{o}M$. The Killing form $\mathcal{B}$ restricted to $\mathfrak{p}$ is
positive definite. Let $\theta$ be the Cartan involution associated with the
previous Cartan decomposition, that is,
$\theta|_{\mathfrak{k}}=\operatorname{id}_{\mathfrak{k}}$, and
$\theta|_{\mathfrak{p}}=-\operatorname{id}_{\mathfrak{p}}$. The equation
$\langle X,Y\rangle=-\mathcal{B}(X,\theta Y)$, $X$, $Y\in\mathfrak{g}$,
defines a positive definite inner product on $\mathfrak{g}$ that will be used
extensively from now on. We normalize the metric on $M$ so that its
restriction to $\mathfrak{p}\times\mathfrak{p}$ is precisely the inner product
defined above.

As a matter of notation, if $\mathfrak{v}$ and $\mathfrak{w}$ are subspaces of
$\mathfrak{g}$ we denote the orthogonal complement of $\mathfrak{w}$ in
$\mathfrak{v}$ as

$\mathfrak{v}\ominus\mathfrak{w}=\\{X\in\mathfrak{v}:\langle
X,Y\rangle=0,\text{ for all $Y\in\mathfrak{w}$}\\}.$

We also denote by $\mathfrak{v}_{\mathfrak{w}}$ the orthogonal projection of
$\mathfrak{v}$ onto $\mathfrak{w}$. The norm of $X\in\mathfrak{g}$ is denoted
by $\lvert X\rvert$.

If $F\colon\mathfrak{g}\to\mathfrak{g}$ is a linear map, its adjoint is the
map $F^{*}\colon\mathfrak{g}\to\mathfrak{g}$ satisfying $\langle
F(X),Y\rangle=\langle X,F^{*}(Y)\rangle$ for any $X$, $Y\in\mathfrak{g}$.
Recall that, for $g\in G$, conjugation by $g$ is denoted by $I_{g}\colon G\to
G$, $h\mapsto ghg^{-1}$. The map $\operatorname{Ad}\colon
G\to\mathsf{GL}(\mathfrak{g})$, $g\mapsto\operatorname{Ad}(g)=I_{g*}$, is
called the adjoint representation of $\mathfrak{g}$, and the differential of
$\operatorname{Ad}$ is the adjoint map
$\operatorname{ad}\colon\mathfrak{g}\to\mathfrak{gl}(\mathfrak{g})$ given by
$\operatorname{ad}(X)(Y)=[X,Y]$. Since $\operatorname{Ad}$ is a homomorphism
of Lie groups and $\operatorname{ad}$ is its differential, they are related by
the Lie exponential map $\operatorname{Exp}\colon\mathfrak{g}\to G$ and the
matrix exponential map of $\mathsf{GL}(\mathfrak{g})$ as
$\operatorname{Ad}(\operatorname{Exp}(X))=e^{\operatorname{ad}(X)}$,
$X\in\mathfrak{g}$. Then, it follows that

$\displaystyle\operatorname{ad}(X)^{*}$
$\displaystyle{}=-\operatorname{ad}(\theta X),$
$\displaystyle\operatorname{Ad}(\operatorname{Exp}(X))^{*}$
$\displaystyle{}=e^{-\operatorname{ad}(\theta X)}.$

Note that $\operatorname{ad}(X)$ is skew-adjoint if $X\in\mathfrak{k}$ and
self-adjoint if $X\in\mathfrak{p}$.

We choose a maximal abelian subspace $\mathfrak{a}$ of $\mathfrak{p}$. Any two
maximal abelian subspaces of $\mathfrak{p}$ are conjugate by an element of
$K$; in particular they have the same dimension, which is called the rank of
$M$, denoted by $r=\dim\mathfrak{a}$. If $\mathfrak{a}^{*}$ denotes the dual
vector space of $\mathfrak{a}$, for each $\lambda\in\mathfrak{a}^{*}$ we
define

$\mathfrak{g}_{\lambda}=\\{X\in\mathfrak{g}:\operatorname{ad}(H)X=\lambda(H)X,\text{
for all $H\in\mathfrak{a}$}\\}.$

If $\lambda\neq 0$ and $\mathfrak{g}_{\lambda}\neq\\{0\\}$ then $\lambda$ is
called a restricted root (or simply root). We denote by $\Sigma$ the set of
restricted roots. Since $\mathfrak{a}$ is abelian,
$\operatorname{ad}(\mathfrak{a})$ is a commuting family of self-adjoint
operators. The corresponding simultaneous diagonalization

$\mathfrak{g}=\mathfrak{g}_{0}\oplus\Bigl{(}\bigoplus_{\lambda\in\Sigma}\mathfrak{g}_{\lambda}\Bigr{)},$

is called the restricted root space decomposition of $\mathfrak{g}$ determined
by $\mathfrak{a}$. Here, we have
$\mathfrak{g}_{0}=\mathfrak{k}_{0}\oplus\mathfrak{a}$, where
$\mathfrak{k}_{0}=\mathfrak{g}_{0}\cap\mathfrak{k}=Z_{\mathfrak{k}}(\mathfrak{a})$
is the centralizer of $\mathfrak{a}$ in $\mathfrak{k}$. Moreover,
$\theta\mathfrak{g}_{\lambda}=\mathfrak{g}_{-\lambda}$ and
$[\mathfrak{g}_{\lambda},\mathfrak{g}_{\mu}]\subseteq\mathfrak{g}_{\lambda+\mu}$
for any $\lambda$, $\mu\in\Sigma$.

We also use the following notation: for $\lambda\in\Sigma$ we denote by
$H_{\lambda}\in\mathfrak{a}$ the metric dual of $\lambda$, which is defined as
$\langle H_{\lambda},H\rangle=\lambda(H)$, $H\in\mathfrak{a}$. The inner
product on $\mathfrak{g}$ induces an inner product on $\mathfrak{a}$ by
restriction, and on $\mathfrak{a}^{*}$ by duality as
$\langle\lambda,\mu\rangle=\langle H_{\lambda},H_{\mu}\rangle$, $\lambda$,
$\mu\in\Sigma$.

The set $\Sigma$ is a (nonreduced) root system. We choose a subset
$\Sigma^{+}$ of positive roots, and $\Lambda$ the corresponding set of simple
roots. Then, the cardinality of $\Lambda$ is equal to the rank of
$\mathfrak{g}$, and $\Lambda$ is a basis of the vector space $\mathfrak{a}$.
We define
$\mathfrak{n}=\bigoplus_{\lambda\in\Sigma^{+}}\mathfrak{g}_{\lambda}$, which
is a nilpotent subalgebra of $\mathfrak{g}$. Then,
$\mathfrak{g}=\mathfrak{k}\oplus\mathfrak{a}\oplus\mathfrak{n}$ is a direct
sum of vector spaces called the Iwasawa decomposition of $\mathfrak{g}$. The
Lie algebra $\mathfrak{a}\oplus\mathfrak{n}$ is solvable and its derived Lie
algebra is $\mathfrak{n}$. If $A$ and $N$ denote the connected subgroups of
$G$ whose Lie algebras are $\mathfrak{a}$ and $\mathfrak{n}$, respectively,
the map $K\times A\times N\to G$, $(k,a,n)\mapsto kan$, is a diffeomorphism.
The corresponding decomposition $G=KAN$ is called the Iwasawa decomposition of
$G$. The connected subgroup $AN$ whose Lie algebra is
$\mathfrak{a}\oplus\mathfrak{n}$ is simply connected, solvable, and acts
simply transitively on $M$. Therefore, $M$ is isometric to the Lie group $AN$
endowed with a left-invariant Riemannian metric. Moreover, the tangent space
$T_{o}M$ can be identified with $\mathfrak{a}\oplus\mathfrak{n}$. The Lie
exponential map $\operatorname{Exp}\colon\mathfrak{a}\oplus\mathfrak{n}\to AN$
is a diffeomorphism.

Since $AN$ acts simply transitively on $M$, we may endow $AN$ with a left
invariant metric $\langle\cdot,\cdot\rangle_{AN}$ such that $M$ and $AN$ are
isometric via the map $g\in AN\mapsto g\cdot o$. The differential of this map
at $e$ is precisely the orthogonal projection of
$\mathfrak{a}\oplus\mathfrak{n}$ onto $\mathfrak{p}$, and the metric in $AN$
satisfies $\langle X,Y\rangle_{AN}=\langle
X_{\mathfrak{a}},Y_{\mathfrak{a}}\rangle+\frac{1}{2}\langle
X_{\mathfrak{n}},Y_{\mathfrak{n}}\rangle$ for every $X$,
$Y\in\mathfrak{a}\oplus\mathfrak{n}$. Furthermore, the Levi-Civita connection
of $AN$ is given in terms of left invariant vector fields as
$4\langle\nabla_{X}Y,Z\rangle_{AN}=\langle[X,Y]+(1-\theta)[\theta
X,Y],Z\rangle$, where $X$, $Y$, $Z\in\mathfrak{a}\oplus\mathfrak{n}$.

Note that for any $X\in\mathfrak{g}$ we have $2X=(1+\theta)X+(1-\theta)X$,
where $(1+\theta)X\in\mathfrak{k}$ and $(1-\theta)X\in\mathfrak{p}$. We use
the following notation in what follows:

$\displaystyle\mathfrak{k}_{\lambda}$
$\displaystyle{}=(1+\theta)\mathfrak{g}_{\lambda}=(1+\theta)\mathfrak{g}_{-\lambda}=\mathfrak{k}\cap(\mathfrak{g}_{\lambda}\oplus\mathfrak{g}_{-\lambda}),$
$\displaystyle\mathfrak{p}_{\lambda}$
$\displaystyle{}=(1-\theta)\mathfrak{g}_{\lambda}=(1-\theta)\mathfrak{g}_{-\lambda}=\mathfrak{p}\cap(\mathfrak{g}_{\lambda}\oplus\mathfrak{g}_{-\lambda}).$

Obviously $\mathfrak{k}_{\lambda}=\mathfrak{k}_{-\lambda}$,
$\mathfrak{p}_{\lambda}=\mathfrak{p}_{-\lambda}$, and
$\mathfrak{k}_{\lambda}\oplus\mathfrak{p}_{\lambda}=\mathfrak{g}_{\lambda}\oplus\mathfrak{g}_{-\lambda}$
for each $\lambda\in\Sigma^{+}$. For $X$, $Y\in\mathfrak{g}_{\lambda}$, we
also use the equality $(1-\theta)[\theta X,Y]=2\langle X,Y\rangle
H_{\lambda}$.

Let $\Lambda=\\{\alpha_{1},\dots,\alpha_{r}\\}$ be the set of simple roots,
and $\\{H^{1},\dots,H^{r}\\}\subseteq\mathfrak{a}$ its dual basis, that is,
$\alpha_{i}(H^{j})=\delta_{i}^{j}$ is the Kronecker delta. We define
$H^{\Lambda}=\sum_{i=1}^{r}H^{i}$. If $\lambda\in\Sigma$, then
$\lambda=\sum_{i=1}^{r}c_{i}\alpha_{i}$, where the $c_{i}$ are integers and
$c_{i}\geq 0$ for all $i\in\\{1,\dots,r\\}$ if $\lambda$ is positive, or
$c_{i}\leq 0$ for all $i$ if $\lambda$ is negative. The integer
$\lambda(H^{\Lambda})=\sum_{i=1}^{r}c_{i}$ is called the level of the root
$\lambda$. This determines a gradation of $\mathfrak{g}$ as

$\mathfrak{g}=\bigoplus_{k\in\mathbb{Z}}\mathfrak{g}^{k},\text{ where
}\mathfrak{g}^{k}=\bigoplus_{\begin{subarray}{c}\lambda\in\Sigma\\\
\lambda(H^{\Lambda})=k\end{subarray}}\mathfrak{g}_{\lambda}.$

Then, $\theta\mathfrak{g}^{k}=\mathfrak{g}^{-k}$, $k\in\mathbb{Z}$, and
$\mathfrak{g}^{0}=\mathfrak{g}_{0}$. According to [14], we have
$\mathfrak{g}^{k+1}=[\mathfrak{g}^{1},\mathfrak{g}^{k}]$,
$\mathfrak{g}^{-k-1}=[\mathfrak{g}^{-1},\mathfrak{g}^{-k}]$, $k\geq 1$. We
also set $\mathfrak{n}^{k}=\mathfrak{g}^{k}$, $k\geq 1$. Thus, $\mathfrak{n}$
is generated by $\mathfrak{n}^{1}$. We also define
$\mathfrak{p}^{k}=\mathfrak{p}\cap(\mathfrak{g}^{k}\oplus\mathfrak{g}^{-k})$.
Finally, there is a highest root of the root space decomposition of
$\mathfrak{g}$. Let $m$ denote its level. In reality we have
$\mathfrak{g}=\bigoplus_{k=-m}^{m}\mathfrak{g}^{k}$,
$\mathfrak{n}=\bigoplus_{k=1}^{m}\mathfrak{n}^{k}$, and
$\mathfrak{p}=\mathfrak{a}\oplus\bigl{(}\bigoplus_{k=1}^{m}\mathfrak{p}^{k}\bigr{)}$.

We define the element

$\delta=\frac{1}{2}\sum_{\lambda\in\Sigma^{+}}(\dim\mathfrak{g}_{\lambda})\lambda.$

If $s_{\alpha}\colon\mathfrak{a}^{*}\to\mathfrak{a}^{*}$ is the root
reflection with respect to a simple root $\alpha\in\Lambda$, it follows from
[15, Theorem 6.57] that $s_{\alpha}$ is induced by an element of
$N_{K}(\mathfrak{a})$, the normalizer of $\mathfrak{a}$ in $K$. In particular,
$\dim\mathfrak{g}_{s_{\alpha}(\lambda)}=\dim\mathfrak{g}_{\lambda}$ for every
$\lambda\in\Sigma$. On the other hand, by [15, Lemma 2.61] $s_{\alpha}$
permutes all positive roots linearly independent from $\alpha$, while sending
$\alpha$ to $-\alpha$. As a consequence,
$s_{\alpha}(\delta)=\delta-(\dim\mathfrak{g}_{\alpha})\alpha-2(\dim\mathfrak{g}_{2\alpha})\alpha$.
Taking the inner product with $\alpha$ yields

(1)
$2\langle\delta,\alpha\rangle=\lvert\alpha\rvert^{2}(\dim\mathfrak{g}_{\alpha}+2\dim\mathfrak{g}_{2\alpha}).$

###  2.2. Parabolic subalgebras and canonical extensions

Let $\Phi$ be a subset of simple roots. We denote by $\Sigma_{\Phi}$ the
subset of $\Sigma$ spanned by $\Phi$, and by $\Sigma_{\Phi}^{+}$ the
corresponding set of positive roots inside $\Sigma_{\Phi}$. Then
$\Sigma_{\Phi}$ is a root subsystem of $\Sigma$ and $\Phi$ is a simple system
for $\Sigma_{\Phi}$. We define

$\displaystyle\mathfrak{a}_{\Phi}$
$\displaystyle{}=\bigcap_{\alpha\in\Phi}\ker\alpha,$
$\displaystyle\mathfrak{l}_{\Phi}$
$\displaystyle{}=\mathfrak{g}_{0}\oplus\bigoplus_{\lambda\in\Sigma_{\Phi}}\mathfrak{g}_{\lambda},$
$\displaystyle\mathfrak{n}_{\Phi}$
$\displaystyle{}=\bigoplus_{\lambda\in\Sigma^{+}\setminus\Sigma_{\Phi}^{+}}\mathfrak{g}_{\lambda}.$

Then, $\mathfrak{a}_{\Phi}$ is abelian, $\mathfrak{l}_{\Phi}$ is reductive,
and $\mathfrak{n}_{\Phi}$ is nilpotent. Moreover,
$[\mathfrak{l}_{\Phi},\mathfrak{n}_{\Phi}]\subseteq\mathfrak{n}_{\Phi}$.

By definition,
$\mathfrak{q}_{\Phi}=\mathfrak{l}_{\Phi}\oplus\mathfrak{n}_{\Phi}$ is called
the parabolic subalgebra of $\mathfrak{g}$ determined by $\Phi$ [6]. Let
$\mathfrak{m}_{\Phi}=\mathfrak{l}_{\Phi}\ominus\mathfrak{a}_{\Phi}$ and
$\mathfrak{g}_{\Phi}=[\mathfrak{l}_{\Phi},\mathfrak{l}_{\Phi}]=[\mathfrak{m}_{\Phi},\mathfrak{m}_{\Phi}]$.
It is known that $\mathfrak{g}_{\Phi}$ is semisimple, and that
$\mathfrak{m}_{\Phi}$ normalizes
$\mathfrak{a}_{\Phi}\oplus\mathfrak{n}_{\Phi}$. The decomposition
$\mathfrak{q}_{\Phi}=\mathfrak{m}_{\Phi}\oplus\mathfrak{a}_{\Phi}\oplus\mathfrak{n}_{\Phi}$
is called the Langlands decomposition of $\mathfrak{q}_{\Phi}$.

We denote by $Q_{\Phi}$, $M_{\Phi}$, $G_{\Phi}$, $A_{\Phi}$, and $N_{\Phi}$
the connected subgroups of $G$ whose Lie algebras are $\mathfrak{q}_{\Phi}$,
$\mathfrak{m}_{\Phi}$, $\mathfrak{g}_{\Phi}$, $\mathfrak{a}_{\Phi}$, and
$\mathfrak{n}_{\Phi}$, respectively. The subgroup $Q_{\Phi}$ is the parabolic
subgroup of $G$ associated with $\Phi$, and
$Q_{\Phi}=M_{\Phi}A_{\Phi}N_{\Phi}$ is the corresponding Langlands
decomposition of $Q_{\Phi}$ at Lie group level.

We define $B_{\Phi}=M_{\Phi}\cdot o=G_{\Phi}\cdot o$. Then, $B_{\Phi}$ is a
totally geodesic submanifold of $M$, and since $G_{\Phi}$ is semisimple, it is
a symmetric space of noncompact type and of rank $\lvert\Phi\rvert$. This
submanifold is called the boundary component associated with $\Phi$.

Let $H_{\Phi}$ be a subgroup of the isometry group of $B_{\Phi}$. Since
$G_{\Phi}$ contains the isometry group of $B_{\Phi}$, it follows that
$H_{\Phi}\subseteq G_{\Phi}\subseteq M_{\Phi}$. As $M_{\Phi}$ normalizes
$A_{\Phi}N_{\Phi}$, $H=H_{\Phi}A_{\Phi}N_{\Phi}$ is a subgroup of $G$ acting
isometrically on $M$, and its Lie algebra is
$\mathfrak{h}_{\Phi}\oplus\mathfrak{a}_{\Phi}\oplus\mathfrak{n}_{\Phi}$. The
action of $H$ on $M$ is called the canonical extension to $M$ of the action of
$H_{\Phi}$ on $B_{\Phi}$ [2].

###  2.3. Maximal solvable subalgebras

We say that a subalgebra $\mathfrak{b}\subseteq\mathfrak{g}$ is a Borel
subalgebra if it is a maximal solvable subalgebra of $\mathfrak{g}$. On the
other hand, a subalgebra $\mathfrak{h}\subseteq\mathfrak{g}$ is a Cartan
subalgebra if its complexification $\mathfrak{h}\otimes\mathbb{C}$ is a Cartan
subalgebra of the complex semisimple Lie algebra
$\mathfrak{g}\otimes\mathbb{C}$. In particular, $\mathfrak{h}$ is abelian [15,
Proposition 2.10]. Note, however, that Cartan subalgebras of real semisimple
Lie algebras are not necessarily conjugate.

Any Cartan subalgebra $\mathfrak{h}$ of $\mathfrak{g}$ is conjugate to a
$\theta$-stable subalgebra [15, Proposition 6.59]. Thus, we can assume that
$\theta\mathfrak{h}=\mathfrak{h}$, which means that $\mathfrak{h}$ splits as a
direct sum $\tilde{\mathfrak{t}}\oplus\tilde{\mathfrak{a}}$, where
$\tilde{\mathfrak{t}}\subseteq\mathfrak{k}$ and
$\tilde{\mathfrak{a}}\subseteq\mathfrak{p}$. Both $\tilde{\mathfrak{t}}$ and
$\tilde{\mathfrak{a}}$ are abelian subspaces of $\mathfrak{g}$. In this case,
$\dim\tilde{\mathfrak{t}}$ is called the compact dimension of $\mathfrak{h}$,
and $\dim\tilde{\mathfrak{a}}$ is called the noncompact dimension of
$\mathfrak{h}$. The subalgebra $\tilde{\mathfrak{t}}$ is called the torus part
of $\mathfrak{h}$, and $\tilde{\mathfrak{a}}$ is called the vector part of
$\mathfrak{h}$. We have that $\tilde{\mathfrak{a}}$ induces a root space
decomposition on $\mathfrak{g}$. Note that, in principle,
$\tilde{\mathfrak{a}}$ does not have to be a maximal abelian subspace of
$\mathfrak{p}$ in this case. Root spaces are defined analogously: for each
$\tilde{\lambda}\in\tilde{\mathfrak{a}}$, let

$\tilde{\mathfrak{g}}_{\tilde{\lambda}}=\\{X\in\mathfrak{g}\colon\operatorname{ad}(H)X=\tilde{\lambda}(H)X\text{
for all $H\in\tilde{\mathfrak{a}}$}\\},$

and define $\tilde{\Sigma}$ to be the set of all
$\tilde{\lambda}\in\tilde{\mathfrak{a}}$ such that $\tilde{\lambda}\neq 0$ and
$\tilde{\mathfrak{g}}_{\tilde{\lambda}}\neq 0$. Since the family
$\operatorname{ad}(\tilde{\mathfrak{a}})$ consists again of commuting self-
adjoint endomorphisms, it follows that
$\mathfrak{g}=\tilde{\mathfrak{g}}_{0}\oplus\bigl{(}\bigoplus_{\tilde{\lambda}\in\tilde{\Sigma}}\tilde{\mathfrak{g}}_{\tilde{\lambda}}\bigr{)}$.
Observe that
$\tilde{\mathfrak{t}}\subseteq\tilde{\mathfrak{g}}_{0}\cap\mathfrak{k}$ since
$\mathfrak{h}$ is abelian.

We now relate the previous decomposition to the root space decomposition
induced by a maximal abelian subspace $\mathfrak{a}\subseteq\mathfrak{p}$
containing $\tilde{\mathfrak{a}}$. Let $\Sigma^{\prime}\subseteq\Sigma$ be the
subset of roots that annihilate $\tilde{\mathfrak{a}}$. We then have the
following equalities:

$\displaystyle\tilde{\mathfrak{a}}$
$\displaystyle{}=\bigcap_{\lambda\in\Sigma^{\prime}}\ker\lambda,$
$\displaystyle\tilde{\mathfrak{g}}_{0}$
$\displaystyle{}=\mathfrak{g}_{0}\oplus\Bigl{(}\bigoplus_{\lambda\in\Sigma^{\prime}}\mathfrak{g}_{\lambda}\Bigr{)},$
$\displaystyle\tilde{\mathfrak{g}}_{\lambda}$
$\displaystyle{}=\bigoplus_{\begin{subarray}{c}\lambda\in\Sigma\\\
\lambda|_{\tilde{\mathfrak{a}}}=\tilde{\lambda}\end{subarray}}\mathfrak{g}_{\lambda}.$

Since $\Sigma^{\prime}$ is an abstract root system in
$(\mathfrak{a}\ominus\tilde{\mathfrak{a}})^{*}$, we may give a notion of
positivity on $\Sigma$ that is compatible with that of $\Sigma^{\prime}$ and
$\tilde{\Sigma}$, that is, $\lambda\in\Sigma$ is positive if and only if
$\lambda\in\Sigma^{\prime}$ is positive or
$\lambda\in\Sigma\setminus\Sigma^{\prime}$ and
$\lambda|_{\tilde{\mathfrak{a}}}\in\tilde{\Sigma}^{+}$.

######  Remark 2.1.

For the sake of completeness we explain here how these notions of positivity
can be made compatible. One can define a notion of positivity on
$\tilde{\Sigma}$ by fixing a regular element
$\tilde{H}_{0}\in\tilde{\mathfrak{a}}$ (that is,
$\tilde{\lambda}(\tilde{H}_{0})\neq 0$ for all
$\tilde{\lambda}\in\tilde{\Sigma}$) and declaring
$\tilde{\lambda}\in\tilde{\Sigma}$ to be positive if
$\tilde{\lambda}(\tilde{H}_{0})>0$. We also take a regular element
$H^{\prime}\in\mathfrak{a}\ominus\tilde{\mathfrak{a}}$ and define
$\lambda\in\Sigma^{\prime}$ to be positive whenever $\lambda(H^{\prime})>0$.
We now define $H_{0}=\tilde{H}_{0}+\varepsilon H^{\prime}$, where
$\varepsilon$ is a positive constant. Note that for every
$\lambda\in\Sigma^{\prime}$, $\lambda(H_{0})=\varepsilon\lambda(H^{\prime})$,
so $\lambda(H_{0})$ and $\lambda(H^{\prime})$ have the same sign. Furthermore,
if $\lambda\in\Sigma\setminus\Sigma^{\prime}$, we have
$\lambda(H_{0})=\lambda|_{\tilde{\mathfrak{a}}}(\tilde{H}_{0})+\varepsilon\lambda(H^{\prime})$.
Since the set of roots is finite, we can choose $\varepsilon>0$ sufficiently
small so that $\lambda(H_{0})$ and
$\lambda|_{\tilde{\mathfrak{a}}}(\tilde{H}_{0})$ have the same sign for all
$\lambda\in\Sigma\setminus\Sigma^{\prime}$.

By [19, Theorem 4.1], any Borel subalgebra $\mathfrak{b}$ of $\mathfrak{g}$ is
of the form
$\mathfrak{b}=\tilde{\mathfrak{t}}\oplus\tilde{\mathfrak{a}}\oplus\tilde{\mathfrak{n}}$
for an adequate choice of a Cartan subalgebra
$\mathfrak{h}=\tilde{\mathfrak{t}}\oplus\tilde{\mathfrak{a}}$, a set of
positive elements $\tilde{\Sigma}^{+}\subseteq\tilde{\Sigma}$, and where
$\tilde{\mathfrak{n}}=\bigoplus_{\tilde{\lambda}\in\tilde{\Sigma}^{+}}\tilde{\mathfrak{g}}_{\tilde{\lambda}}$.
We aim to restate this description of $\mathfrak{b}$ directly in terms of the
root system induced by $\mathfrak{a}$.

We consider the subset $\Phi\subseteq\Sigma^{\prime}$ of simple roots
associated with the positivity criterion in $\Sigma^{\prime}$. Note that
$\Phi\subseteq\Lambda$. Indeed, by the construction of our set of positive
roots in $\Sigma$, we have $\Phi\subseteq\Sigma^{+}$. Suppose $\alpha\in\Phi$
is not simple, so that $\alpha=\beta+\gamma$ for two positive roots $\beta$,
$\gamma\in\Sigma^{+}$. Since $\alpha$ is simple in $\Sigma^{\prime}$, we have
that $\beta$ and $\gamma$ cannot be simultaneously in $\Sigma^{\prime}$, and
combining this with the equation $0=\alpha(H_{0})=\beta(H_{0})+\gamma(H_{0})$,
we deduce that neither $\beta$ nor $\gamma$ are in $\Sigma^{\prime}$, and
either $\beta$ or $\gamma$ is negative, a contradiction.

To summarize, we have found a subset $\Phi\subseteq\Lambda$ of simple roots
for which $\Sigma^{\prime}=\Sigma_{\Phi}$ is the root system generated by
$\Phi$ and the following identities hold:
$\tilde{\mathfrak{a}}=\mathfrak{a}_{\Phi}$,
$\tilde{\mathfrak{g}}_{0}=\mathfrak{l}_{\Phi}$ and
$\tilde{\mathfrak{n}}=\mathfrak{n}_{\Phi}$.

We have thus arrived at the following result.

######  Theorem 2.2.

Let $\mathfrak{g}$ be a real semisimple Lie algebra and $\mathfrak{b}$ a Borel
subalgebra of $\mathfrak{g}$. Then $\mathfrak{b}$ contains a Cartan subalgebra
$\mathfrak{h}$. Furthermore, there exists a Cartan decomposition
$\mathfrak{g}=\mathfrak{k}\oplus\mathfrak{p}$, a maximal abelian subspace
$\mathfrak{a}\subseteq\mathfrak{p}$, a choice of simple roots
$\Lambda\subseteq\Sigma$, and a set $\Phi\subseteq\Lambda$ such that
$\mathfrak{b}=\tilde{\mathfrak{t}}\oplus\mathfrak{a}_{\Phi}\oplus\mathfrak{n}_{\Phi}$,
where $\tilde{\mathfrak{t}}$ is an abelian subspace of
$\mathfrak{k}_{\Phi}=\mathfrak{k}\cap\mathfrak{l}_{\Phi}=\mathfrak{k}_{0}\oplus\bigl{(}\bigoplus_{\lambda\in\Sigma_{\Phi}^{+}}\mathfrak{k}_{\lambda}\bigr{)}$.

We say that a Cartan subalgebra $\mathfrak{h}$ (resp. Borel subalgebra
$\mathfrak{b}$) is maximally compact if its compact dimension is maximal, and
maximally noncompact if its noncompact dimension is maximal. Since
$\tilde{\mathfrak{a}}$ is abelian in $\mathfrak{p}$, we have that
$\mathfrak{h}$ (resp. $\mathfrak{b}$) is maximally noncompact if and only if
$\tilde{\mathfrak{a}}$ is a maximal abelian subspace of $\mathfrak{p}$ [15,
Proposition 6.47].

If a Borel subalgebra corresponds to a maximally noncompact Cartan subalgebra,
then $\Phi=\emptyset$ is the empty set,
$\tilde{\mathfrak{a}}={\mathfrak{a}_{\emptyset}}=\mathfrak{a}$ is a maximal
abelian subspace of $\mathfrak{p}$, and $\mathfrak{t}=\tilde{\mathfrak{t}}$ is
a maximal abelian subspace of $\mathfrak{k}_{0}$ [15, Proposition 6.47 and
Lemma 6.62]. This implies ${\mathfrak{n}_{\emptyset}}=\mathfrak{n}$, and thus,
$\mathfrak{b}=\mathfrak{t}\oplus\mathfrak{a}\oplus\mathfrak{n}$.

###  2.4. Polar actions

Let $M$ be a Riemannian manifold and $H$ a connected Lie group acting on $M$
isometrically. Given any point $p\in M$, we denote by $H_{p}$ the isotropy
subgroup at $p$, and by $H\cdot p$ the $H$-orbit through $p$. We assume that
the action of $H$ on $M$ is proper, that is, the map $H\times M\to M\times M$,
$(g,p)\mapsto(p,g\cdot p)$ is proper. In this case, isotropy groups are
compact, the orbit space is Hausdorff, and the orbits of the action are closed
and embedded submanifolds. If the action is also effective, that is, the only
element of the group acting trivially is the identity element, then $H$ can be
assumed to be a closed subgroup of the isometry group of $M$ acting on $M$ in
the natural way. If $p\in M$ is any point, we define the isotropy
representation of $H$ at $p$ as the map $g\in H_{p}\mapsto
g_{*p}\in\mathsf{O}(T_{p}M)$, and the slice representation of $H$ at $p$ as
its restriction to $\mathsf{O}(\nu_{p}(H\cdot p))$. Here, the notation
$\nu_{p}N$ refers to the normal subspace of the submanifold $N\subseteq M$ at
$p$.

Two isometric actions on $M$ are said to be orbit equivalent if there is an
isometry of $M$ that maps the orbits of one action to the orbits of the other
action. They are said to be conjugate if there is an isometry of $M$ that is
equivariant with respect to both actions.

We start by mentioning the following result, which is a refinement of [8,
Lemma 2.5].

######  Lemma 2.3.

Let $M$ be a complete Riemannian manifold and $H$ and $\tilde{H}$ be
connected, not necessarily closed, subgroups of the isometry group of $M$ such
that $H\subseteq\tilde{H}$. Suppose that there exists $o\in M$ such that
$H\cdot o=\tilde{H}\cdot o$ is a closed subset of $M$, and the slice
representation of $\tilde{H}$ at $o$ is trivial. Then $H$ and $\tilde{H}$ act
with the same orbits.

###### Proof.

Let $p\in M$ be arbitrary. Since $H\cdot o$ is closed in $M$, we may find a
point $q\in H\cdot o$ such that the distance from $q$ to $p$ is minimum among
all points of $H\cdot o$. The first variation formula implies that the
minimizing geodesic joining $q$ and $p$ must leave $H\cdot o$ perpendicularly.
Thus, by homogeneity we may assume that $q=o$ and $p=\exp_{o}(\xi_{o})$, with
$\xi_{o}\in\nu_{o}(H\cdot o)$. Let $\xi\in\Gamma(\nu(H\cdot o))$ be the unique
$\tilde{H}$-equivariant vector field whose value at $o$ is $\xi_{o}$ (which
exists because $\xi_{o}$ is fixed by the slice representation of $\tilde{H}$).
Since $H\subseteq\tilde{H}$, $\xi$ is also the unique $H$-equivariant normal
vector field along $H\cdot o$ generated by $\xi_{o}$. Due to [1, Section
2.1.8], we obtain $H\cdot p=\\{\exp_{x}(\xi_{x})\colon x\in H\cdot
o\\}=\tilde{H}\cdot p$. ∎

If $H$ is a closed subgroup of the isometry group of $M$, we say that the
action of $H$ on $M$ is _polar_ if there exists a submanifold $\Sigma$ in $M$
such that:

  1. (i)

$\Sigma$ intersects all the orbits of $H$, and

  2. (ii)

if $p\in\Sigma$, then $T_{p}\Sigma$ and $T_{p}(H\cdot p)$ are orthogonal.

The submanifold $\Sigma$ is called a _section_. Sections are known to be
totally geodesic (see for example [18]). If the section of a polar action is
flat, then the action is called _hyperpolar_.

In this paper we assume that the action of $H$ on $M$ induces a foliation. We
say that $M$ is a Hadamard manifold if it is a simply connected complete
Riemannian manifold with nonpositive sectional curvature. Riemannian symmetric
spaces of noncompact type are examples of Hadamard manifolds. If follows from
[2, Proposition 2.1] that all the orbits of $H$ are principal, that is, all
isotropy groups are conjugate in $H$.

From [2, Theorem 4.1] we have the following criterion of polarity:

######  Proposition 2.4.

Let $M=G/K$ be a Riemannian symmetric space of noncompact type with Cartan
decomposition $\mathfrak{g}=\mathfrak{k}\oplus\mathfrak{p}$. Let $H\subseteq
G$ be a closed subgroup acting on $M$ in such a way that its orbits form a
foliation on $M$. We define

$\mathfrak{h}_{\mathfrak{p}}^{\perp}=\\{\xi\in\mathfrak{p}:\langle\xi,X\rangle=0,\text{
for all $X\in\mathfrak{h}$}\\}.$

Then:

  1. (i)

The group $H$ acts polarly on $M$ if and only if
$\mathfrak{h}_{\mathfrak{p}}^{\perp}$ is a Lie triple system in $\mathfrak{p}$
and
$[\mathfrak{h}_{\mathfrak{p}}^{\perp},\mathfrak{h}_{\mathfrak{p}}^{\perp}]$ is
orthogonal to $\mathfrak{h}$.

  2. (ii)

The group $H$ acts hyperpolarly on $M$ if and only if
$\mathfrak{h}_{\mathfrak{p}}^{\perp}$ is an abelian subspace of
$\mathfrak{p}$.

Moreover, if $H_{\mathfrak{p}}^{\perp}$ denotes the subgroup of $G$ whose Lie
algebra is
$[\mathfrak{h}_{\mathfrak{p}}^{\perp},\mathfrak{h}_{\mathfrak{p}}^{\perp}]\oplus\mathfrak{h}_{\mathfrak{p}}^{\perp}$,
then $H_{\mathfrak{p}}^{\perp}\cdot o$ is a section of the action of $H$ on
$M$.

The first criterion of polarity is credited to Gorodski [12].

Recall that a subspace $\mathfrak{v}$ of $\mathfrak{p}$ is called a Lie triple
system if $[\mathfrak{v},[\mathfrak{v},\mathfrak{v}]]\subseteq\mathfrak{v}$.
In this case $[\mathfrak{v},\mathfrak{v}]\oplus\mathfrak{v}$ is a
$\theta$-stable subalgebra of $\mathfrak{g}$ and the orbit through $o\in M$ of
the connected subgroup of $G$ whose Lie algebra is
$[\mathfrak{v},\mathfrak{v}]\oplus\mathfrak{v}$ is a totally geodesic
submanifold of $M$.

##  3\. Examples of homogeneous polar foliations

We now introduce the two families of polar homogeneous foliations on $M=G/K$
whose section is homothetic to the hyperbolic plane, and describe their
extrinsic geometry. We assume the notation used in Section 2.

######  Theorem 3.1.

Let $M=G/K$ be a symmetric space of noncompact type and choose an Iwasawa
decomposition $\mathfrak{g}=\mathfrak{k}\oplus\mathfrak{a}\oplus\mathfrak{n}$
of $\mathfrak{g}$. Let $\alpha\in\Lambda$ be a simple root, and consider the
following subspaces of $\mathfrak{a}\oplus\mathfrak{n}$:

  1. (i)

$\mathfrak{s}_{\xi}=(\mathfrak{a}\ominus\mathbb{R}H_{\alpha})\oplus(\mathfrak{n}\ominus\mathbb{R}\xi)$,
where $\xi\in\mathfrak{g}_{\alpha}$ is a nonzero vector.

  2. (ii)

$\mathfrak{s}_{\mathfrak{v}}=\mathfrak{a}\oplus(\mathfrak{n}\ominus\mathfrak{v})$,
where $\mathfrak{v}\subseteq\mathfrak{g}_{\alpha}$ is an abelian plane inside
$\mathfrak{g}_{\alpha}$.

The subspaces $\mathfrak{s}_{\xi}$ and $\mathfrak{s}_{\mathfrak{v}}$ are Lie
subalgebras of $\mathfrak{a}\oplus\mathfrak{n}$. The corresponding connected
subgroups $S_{\xi}$, $S_{\mathfrak{v}}$ act polarly on $M$ inducing a
codimension two foliation whose section is a totally geodesic
$\mathbb{R}\mathsf{H}^{2}$ with constant curvature $-\lvert\alpha\rvert^{2}$.

###### Proof.

It is clear that $\mathfrak{s}_{\xi}$ and $\mathfrak{s}_{\mathfrak{v}}$ are
subalgebras of $\mathfrak{a}\oplus\mathfrak{n}$, so we can consider the
connected Lie subgroups $S_{\xi}$, $S_{\mathfrak{v}}$ associated with these
subalgebras. Since $AN$ acts simply transitively on $M$ and
$\operatorname{Exp}\colon\mathfrak{a}\oplus\mathfrak{n}\to AN$ is a
diffeomorphism, it follows that $S_{\xi}$ and $S_{\mathfrak{v}}$ are closed
subgroups inducing a homogeneous foliation on $M$ of codimension $2$. It
remains to show that both subgroups act polarly.

Firstly, if $S=S_{\xi}$, we see that
$\mathfrak{s}_{\mathfrak{p}}^{\perp}=\operatorname{span}\\{H_{\alpha},(1-\theta)\xi\\}$.
Now, a direct computation shows that
$[\mathfrak{s}_{\mathfrak{p}}^{\perp},\mathfrak{s}_{\mathfrak{p}}^{\perp}]=\mathbb{R}(1+\theta)\xi$
is orthogonal to $\mathfrak{s}$, and
$[\mathfrak{s}_{\mathfrak{p}}^{\perp},[\mathfrak{s}_{\mathfrak{p}}^{\perp},\mathfrak{s}_{\mathfrak{p}}^{\perp}]]$
is spanned by $[H_{\alpha},(1+\theta)\xi]=\lvert\alpha\rvert^{2}(1-\theta)\xi$
and
$[(1-\theta)\xi,(1+\theta)\xi]=-(1-\theta)[\theta\xi,\xi]=-2\lvert\xi\rvert^{2}H_{\alpha}$.
We therefore obtain
$[\mathfrak{s}_{\mathfrak{p}}^{\perp},[\mathfrak{s}_{\mathfrak{p}}^{\perp},\mathfrak{s}_{\mathfrak{p}}^{\perp}]]=\mathfrak{s}_{\mathfrak{p}}^{\perp}$,
which means that $\mathfrak{s}_{\mathfrak{p}}^{\perp}$ is a Lie triple system.
By applying Proposition 2.4, we deduce that $S_{\xi}$ acts polarly, as
desired. Note that if $S_{\mathfrak{p}}^{\perp}\cdot o$ is the section through
$o$, then $S_{\mathfrak{p}}^{\perp}\cdot o$ is a closed, simply connected,
totally geodesic surface whose tangent space is
$\mathfrak{s}_{\mathfrak{p}}^{\perp}$. Its sectional curvature can be
calculated using for example [13, Chapter IV, Theorem 4.2], which yields

$\displaystyle\operatorname{sec}(S_{\mathfrak{p}}^{\perp}\cdot o)$
$\displaystyle{}=\frac{-\langle[[H_{\alpha},(1-\theta)\xi],(1-\theta)\xi],H_{\alpha}\rangle}{\lvert(1-\theta)\xi\rvert^{2}\lvert
H_{\alpha}\rvert^{2}}$
$\displaystyle{}=\frac{-\lvert[H_{\alpha},(1-\theta)\xi]\rvert^{2}}{2\lvert\xi\rvert^{2}\lvert\alpha\rvert^{2}}=\frac{-\lvert(1+\theta)\lvert\alpha\rvert^{2}\xi\rvert^{2}}{{2\lvert\xi\rvert^{2}\lvert\alpha\rvert^{2}}}=-\lvert\alpha\rvert^{2},$

so $S_{\mathfrak{p}}^{\perp}\cdot o$ is a real space form of constant
curvature $-\lvert\alpha\rvert^{2}$.

In the case $S=S_{\mathfrak{v}}$, the normal space is
$\mathfrak{s}_{\mathfrak{p}}^{\perp}=(1-\theta)\mathfrak{v}$. Choose two
orthogonal vectors $\xi$, $\eta\in\mathfrak{v}$ with norm ${1}/{\sqrt{2}}$.
Since $[(1-\theta)\xi,(1-\theta)\eta]=-2[\xi,\theta\eta]\in\mathfrak{k}_{0}$,
it follows that
$[\mathfrak{s}_{\mathfrak{p}}^{\perp},\mathfrak{s}_{\mathfrak{p}}^{\perp}]=\mathbb{R}[\xi,\theta\eta]$
is perpendicular to $\mathfrak{s}$. Furthermore,
$\theta[\xi,\theta\eta]=[\xi,\theta\eta]$ yields

$\displaystyle[(1-\theta)\xi,[\xi,\theta\eta]]$
$\displaystyle{}=(1-\theta)[\xi,[\theta\xi,\eta]]=-(1-\theta)[\eta,[\xi,\theta\xi]]$
$\displaystyle{}=(1-\theta)[\eta,\lvert\xi\rvert^{2}H_{\alpha}]=-\lvert\xi\rvert^{2}\lvert\alpha\rvert^{2}(1-\theta)\eta\in\mathfrak{s}_{\mathfrak{p}}^{\perp},$

and a similar calculation gives
$[(1-\theta)\eta,[\xi,\theta\eta]]=\lvert\alpha\rvert^{2}\lvert\eta\rvert^{2}(1-\theta)\xi\in\mathfrak{s}_{\mathfrak{p}}^{\perp}$,
and thus
$[\mathfrak{s}_{\mathfrak{p}}^{\perp},[\mathfrak{s}_{\mathfrak{p}}^{\perp},\mathfrak{s}_{\mathfrak{p}}^{\perp}]]=\mathfrak{s}_{\mathfrak{p}}^{\perp}$.
Proposition 2.4 readily implies that the action of $S_{\mathfrak{v}}$ is polar
with section $S_{\mathfrak{p}}^{\perp}\cdot o$. The same argument given in the
previous paragraph allows us to determine the section by computing its
curvature. In this case, taking into account our previous calculations,

$\operatorname{sec}(S_{\mathfrak{p}}^{\perp}\cdot
o)=-\frac{\langle[[(1-\theta)\xi,(1-\theta)\eta],(1-\theta)\eta],(1-\theta)\xi\rangle}{\lvert(1-\theta)\xi\rvert^{2}\lvert(1-\theta)\eta\rvert^{2}}=-\lvert\alpha\rvert^{2},$

which finishes the proof. ∎

The previous theorem shows that the examples that appear in Theorem B give
rise to homogeneous polar foliations. Furthermore, it follows from the next
lemma that different choices of $\xi$ in case (i) or of $\mathfrak{v}$ in (ii)
given orbit equivalent actions.

######  Lemma 3.2.

Let $\alpha\in\Lambda$ and $k\geq 1$. Then, the group $K_{0}$ acts
transitively on the set of abelian subspaces of dimension $k$ of
$\mathfrak{g}_{\alpha}$.

###### Proof.

Following [13, Chapter IX, §2], we consider the Lie subalgebra
$\mathfrak{g}^{\alpha}$ generated by $\mathfrak{g}_{\alpha}$ and
$\mathfrak{g}_{-\alpha}$. This Lie algebra is simple and its Cartan
decomposition is
$\mathfrak{g}^{\alpha}=\mathfrak{k}^{\alpha}\oplus\mathfrak{p}^{\alpha}$, with
$\mathfrak{k}^{\alpha}=\mathfrak{k}\cap\mathfrak{g}^{\alpha}$,
$\mathfrak{p}^{\alpha}=\mathfrak{p}\cap\mathfrak{g}^{\alpha}$. It turns out
that $\mathbb{R}H_{\alpha}$ is a maximal abelian subspace of
$\mathfrak{p}^{\alpha}$, and the root space decomposition of
$\mathfrak{g}^{\alpha}$ is
$\mathfrak{g}^{\alpha}=\mathfrak{g}_{-2\alpha}\oplus\mathfrak{g}_{-\alpha}\oplus(\mathfrak{k}_{0}^{\alpha}\oplus\mathbb{R}H_{\alpha})\oplus\mathfrak{g}_{\alpha}\oplus\mathfrak{g}_{2\alpha}$,
where $\mathfrak{k}_{0}^{\alpha}$ is the centralizer of $\mathbb{R}H_{\alpha}$
in $\mathfrak{k}^{\alpha}$, and
$\mathfrak{k}_{0}^{\alpha}=\mathfrak{k}_{0}\cap\mathfrak{g}^{\alpha}$. Let
$G^{\alpha}$, $K^{\alpha}$, $K_{0}^{\alpha}$ be the connected subgroups of $G$
whose Lie algebras are $\mathfrak{g}^{\alpha}$, $\mathfrak{k}^{\alpha}$, and
$\mathfrak{k}_{0}^{\alpha}$. Then, by [13, Chapter IX, Lemma 2.3], we have
$K^{\alpha}=K\cap G^{\alpha}$ and $K_{0}^{\alpha}=K_{0}\cap G^{\alpha}$.
Therefore, in order to prove this lemma, it suffices to show that
$K_{0}^{\alpha}$ acts transitively on the set of abelian subspaces of
$\mathfrak{g}_{\alpha}$.

Obviously, $G^{\alpha}/K^{\alpha}$ is a Riemannian symmetric space of
noncompact type and rank one, that is, a hyperbolic space
$\mathbb{F}\mathsf{H}^{n}$, where
$\mathbb{F}\in\\{\mathbb{R},\mathbb{C},\mathbb{H},\mathbb{O}\\}$ and $n\geq 2$
($n=2$ if $\mathbb{F}=\mathbb{O}$). Note that
$\mathfrak{g}_{\alpha}\cong\mathbb{F}^{n-1}$ and
$\dim\mathfrak{g}_{2\alpha}=\dim_{\mathbb{R}}\mathbb{F}-1$. If
$\mathbb{F}=\mathbb{R}$, then $\mathfrak{g}_{\alpha}$ is abelian and
$K_{0}^{\alpha}\cong\mathsf{SO}(n-1)$ acts in the standard way on
$\mathfrak{g}_{\alpha}$; this action is transitive on the Grassmannian of
$k$-planes of $\mathbb{R}^{n-1}$. If $\mathbb{F}=\mathbb{O}$, then the only
nonzero abelian subspaces of $\mathfrak{g}_{\alpha}\cong\mathbb{O}$ are
1-dimensional, and $K_{0}^{\alpha}\cong\mathsf{Spin}(7)$ acts on $\mathbb{O}$
by its irreducible 8-dimensional spin representation, which is transitive in
$\mathsf{S}^{7}$ [5]. Finally, if $\mathbb{F}\in\\{\mathbb{C},\mathbb{H}\\}$,
recall that abelian subspaces of $\mathfrak{g}_{\alpha}$ are precisely totally
real subspaces of $\mathfrak{g}_{\alpha}\cong\mathbb{F}^{n-1}$. In these cases
we have the standard action of $\mathsf{S}(\mathsf{U}(n-1)\mathsf{U}(1))$ on
$\mathbb{C}^{n-1}$ if $\mathbb{F}=\mathbb{C}$, and the standard action of
$\mathsf{Sp}(n-1)\mathsf{Sp}(1)$ on $\mathbb{H}^{n-1}$ if
$\mathbb{F}=\mathbb{H}$. Thus, if $\mathfrak{v}_{1}$ and $\mathfrak{v}_{2}$
are two totally real subspaces of $\mathfrak{g}_{\alpha}$ of dimension $k$,
choose an orthogonal basis of $\mathfrak{v}_{1}$ and an orthogonal basis of
$\mathfrak{v}_{2}$. Since $\mathfrak{v}_{1}$ and $\mathfrak{v}_{2}$ are
totally real, these two bases are not only orthogonal, but
$\mathbb{F}$-orthogonal. By definition of $\mathsf{U}(n-1)$ or
$\mathsf{Sp}(n-1)$ it is then clear that there is an element of
$K_{0}^{\alpha}$ that maps one basis to the other. This finishes the proof. ∎

We now exhibit examples (i) and (ii) of Theorem B as canonical extensions of
actions on a rank one boundary component. Let $\alpha\in\Lambda$ be a simple
root, $\xi\in\mathfrak{g}_{\alpha}$ a unit vector and
$\mathfrak{v}\subseteq\mathfrak{g}_{\alpha}$ an abelian plane. We consider the
set $\Phi=\\{\alpha\\}\subseteq\Lambda$. Then, the subalgebras constructed in
Section 2.2 take the form

$\begin{split}\mathfrak{l}_{\Phi}={}&\mathfrak{g}_{0}\oplus\mathfrak{g}_{-2\alpha}\oplus\mathfrak{g}_{-\alpha}\oplus\mathfrak{g}_{\alpha}\oplus\mathfrak{g}_{2\alpha},\\\
\mathfrak{a}_{\Phi}={}&\ker\alpha,\\\
\mathfrak{n}_{\Phi}={}&\mathfrak{n}\ominus(\mathfrak{g}_{\alpha}\oplus\mathfrak{g}_{2\alpha}),\\\
\mathfrak{m}_{\Phi}={}&\mathfrak{k}_{0}\oplus\mathbb{R}H_{\alpha}\oplus\mathfrak{g}_{-2\alpha}\oplus\mathfrak{g}_{-\alpha}\oplus\mathfrak{g}_{\alpha}\oplus\mathfrak{g}_{2\alpha},\end{split}$

and $B_{\Phi}=M_{\Phi}\cdot o$ is a rank one noncompact symmetric space whose
tangent space at $o$ is
$T_{o}B_{\Phi}=\mathbb{R}H_{\alpha}\oplus\mathfrak{p}_{\alpha}\oplus\mathfrak{p}_{2\alpha}$.
Consider the subalgebras
$\hat{\mathfrak{s}}_{\xi}=(\mathfrak{g}_{\alpha}\ominus\mathbb{R}\xi)\oplus\mathfrak{g}_{2\alpha}$
and
$\hat{\mathfrak{s}}_{\mathfrak{v}}=\mathbb{R}H_{\alpha}\oplus(\mathfrak{g}_{\alpha}\ominus\mathfrak{v})\oplus\mathfrak{g}_{2\alpha}$
of $\mathfrak{g}_{\Phi}=[\mathfrak{m}_{\Phi},\mathfrak{m}_{\Phi}]$. The
corresponding connected subgroups $\hat{S}_{\xi}$ and $\hat{S}_{\mathfrak{v}}$
act polarly on $B_{\Phi}$ inducing a foliation, due to Theorem 3.1. Recall
that the canonical extensions of the actions of $\hat{S}_{\xi}$ and
$\hat{S}_{\mathfrak{v}}$ are the actions of the subgroups
$\hat{S}_{\xi}A_{\Phi}N_{\Phi}$ and $\hat{S}_{\mathfrak{v}}A_{\Phi}N_{\Phi}$,
respectively. Observe that
$\hat{\mathfrak{s}}_{\xi}\oplus\mathfrak{a}_{\Phi}\oplus\mathfrak{n}_{\Phi}=\mathfrak{s}_{\xi}$,
while
$\hat{\mathfrak{s}}_{\mathfrak{v}}\oplus\mathfrak{a}_{\Phi}\oplus\mathfrak{n}_{\Phi}=\mathfrak{s}_{\mathfrak{v}}$,
and this readily implies that these canonical extensions are precisely the
actions of $S_{\xi}$ and $S_{\mathfrak{v}}$. We deduce that Theorem B implies
Theorem A.

The remaining part of this section will be devoted to computing the mean
curvature of the orbits in each example. To this end, we consider the solvable
model $M=AN$ discussed in Section 2.1. If
$\mathfrak{s}\subseteq\mathfrak{a}\oplus\mathfrak{n}$, we refer to its
orthogonal complement in $\mathfrak{a}\oplus\mathfrak{n}$ with respect to
$\langle\cdot,\cdot\rangle_{AN}$ as $\mathfrak{s}^{\perp}$. Note that if
$S\subseteq AN$ is a Lie subgroup, the isometry $g\in AN\mapsto g\cdot o\in M$
induces an orbit equivalence between the action of $S$ on $M$ and the action
of $S$ on $AN$ by left multiplication.

Recall that if $M\subseteq\widetilde{M}$ is a submanifold of a Riemannian
manifold with second fundamental form $\mathbb{II}$, we define the mean
curvature vector of $M$ at $p$ as
$\mathcal{H}_{p}=\sum_{i}\mathbb{II}(e_{i},e_{i})$, where $\\{e_{i}\\}_{i}$ is
an orthonormal basis of $T_{p}M$. In other words, $\mathcal{H}$ is the trace
of the second fundamental form. If $S\subseteq AN$ is a connected subgroup of
$AN$, it is easy to see from the formula for the Levi-Civita connection that
the second fundamental form of $S\subseteq AN$ at $e$ satisfies the identity

(2)
$\langle\mathbb{II}(X,X),\eta\rangle_{AN}=\frac{1}{4}\langle(1-\theta)[\theta
X,X],\eta\rangle$

for each $X\in\mathfrak{s}$ and $\eta\in\mathfrak{s}^{\perp}$.

Let us start by discussing foliations of type (i).

######  Proposition 3.3.

Let $\alpha\in\Lambda$ and $\xi\in\mathfrak{g}_{\alpha}$ a vector such that
$\langle\xi,\xi\rangle=1$. All the orbits of $S_{\xi}$ are isometrically
congruent. Furthermore, the mean curvature vector of $S_{\xi}$ at $e$ is given
by the following expression:

$\mathcal{H}_{e}=\left(\dim\mathfrak{g}_{\alpha}+2\dim\mathfrak{g}_{2\alpha}-1\right)H_{\alpha}.$

###### Proof.

Observe that
$\mathfrak{s}_{\xi}=\ker\alpha\oplus(\mathfrak{n}\ominus\mathbb{R}\xi)$ is an
ideal of $\mathfrak{a}\oplus\mathfrak{n}$. As a consequence, if $g\in AN$ is
an arbitrary point, we have $S_{\xi}\cdot g=gg^{-1}S_{\xi}g=gS_{\xi}$, because
$S_{\xi}$ is a normal subgroup of $AN$. Thus, $S_{\xi}\cdot g$ is isometric to
$S_{\xi}$ via the left multiplication by $g$.

We proceed to compute $\mathcal{H}_{o}$. For this, it suffices to determine
the vectors $\mathbb{II}(H,H)$ and $\mathbb{II}(X,X)$ for each
$H\in\mathfrak{a}$ and $X\in\mathfrak{g}_{\lambda}\ominus\mathbb{R}\xi$, where
$\lambda$ is any positive root. Given any $H\in\mathfrak{a}$, it is clear from
(2) that $\mathbb{II}(H,H)=0$. On the other hand, if $\lambda\in\Sigma^{+}$
and $X\in\mathfrak{g}_{\lambda}$ is a unit vector orthogonal to $\xi$, we
obtain that $1=\langle X,X\rangle_{AN}=|X|^{2}/2$ and
$\langle\mathbb{II}(X,X),\eta\rangle_{AN}=\frac{1}{2}\langle|X|^{2}H_{\lambda},\eta\rangle=\langle
H_{\lambda},\eta\rangle=\langle\frac{\langle\lambda,\alpha\rangle}{|\alpha|^{2}}H_{\alpha},\eta\rangle_{AN}$
for each
$\eta\in\mathfrak{s}^{\perp}=\operatorname{span}\\{H_{\alpha},\xi\\}$, which
means that
$\mathbb{II}(X,X)=\frac{\langle\lambda,\alpha\rangle}{|\alpha|^{2}}H_{\alpha}$.
In conclusion,

$\displaystyle\mathcal{H}_{e}={}$
$\displaystyle\frac{1}{|\alpha|^{2}}\bigg{(}\sum_{\lambda\in\Sigma^{+}\setminus\\{\alpha\\}}(\dim\mathfrak{g}_{\lambda})\langle\lambda,\alpha\rangle+(\dim\mathfrak{g}_{\alpha}-1)|\alpha|^{2}\bigg{)}H_{\alpha}=\frac{1}{|\alpha|^{2}}(\langle
2\delta,\alpha\rangle-|\alpha|^{2})H_{\alpha}$ $\displaystyle={}$
$\displaystyle(\dim\mathfrak{g}_{\alpha}+2\dim\mathfrak{g}_{2\alpha}-1)H_{\alpha},$

where we have used (1) for the last equality. ∎

A direct consequence of the previous proposition is that the foliation induced
by $S_{\xi}$ consists of congruent minimal submanifolds if and only if
$2\alpha\notin\Sigma$ and $\dim\mathfrak{g}_{\alpha}=1$. If this is the case,
then $\mathfrak{s}_{\xi}=\mathfrak{a}_{\Phi}\oplus\mathfrak{n}_{\Phi}$ for
$\Phi=\\{\alpha\\}$, and $B_{\Phi}$ is homothetic to the hyperbolic plane.
Furthermore, $\hat{\mathfrak{s}}_{\xi}=0$, which means that $\hat{S}_{\xi}$
acts trivially on $B_{\Phi}$. More generally, the extrinsic geometry of the
orbits for the solvable part $A_{\Phi}N_{\Phi}$ of a parabolic subgroup was
studied by Tamaru in [24], where he proved that for any choice of
$\Phi\subseteq\Lambda$ the subgroup $A_{\Phi}N_{\Phi}$ induces a harmonic
foliation on $M$ whose orbits are congruent.

We now consider the foliations from case (ii). In this setting, the orbits are
not isometrically congruent, as their mean curvature does not have constant
length. More precisely:

######  Proposition 3.4.

Let $\alpha\in\Lambda$ be a simple root and
$\mathfrak{v}\subseteq\mathfrak{g}_{\alpha}$ an abelian subspace of dimension
$2$. Fix a vector $\xi\in\mathfrak{v}$ with $|\xi|=1$, and denote by
$\mathcal{H}_{t}$ the mean curvature vector of
$S_{\mathfrak{v}}\cdot\operatorname{Exp}(t\xi)$ at $\operatorname{Exp}(t\xi)$
and by $L_{\operatorname{Exp}(t\xi)}\colon AN\to AN$ the left translation by
$\operatorname{Exp}(t\xi)$. Then,

$(L_{\operatorname{Exp}(-t\xi)})_{*\operatorname{Exp}(t\xi)}\mathcal{H}_{t}=\frac{t|\alpha|^{2}}{2+t^{2}|\alpha|^{2}}(\dim\mathfrak{g}_{\alpha}+2\dim\mathfrak{g}_{2\alpha}-1)(tH_{\alpha}-2\xi).$

In particular, the orbit through $\operatorname{Exp}(t\xi)$ is minimal if and
only if $t=0$.

###### Proof.

Firstly, if $g=\operatorname{Exp}(t\xi)\in AN$, we deduce that
$S_{\mathfrak{v}}\cdot g=gg^{-1}S_{\mathfrak{v}}g=g(g^{-1}S_{\mathfrak{v}}g)$
is isometric to $g^{-1}S_{\mathfrak{v}}g$ by left translation. Thus, it
suffices to compute the mean curvature $\tilde{\mathcal{H}}_{t}$ of
$\tilde{S}=g^{-1}S_{\mathfrak{v}}g$ at $e$. To this end, we compute the Lie
algebra
$\tilde{\mathfrak{s}}=\operatorname{Ad}(g^{-1})\mathfrak{s}_{\mathfrak{v}}$ of
$g^{-1}S_{\mathfrak{v}}g$. Observe that
$\operatorname{Ad}(g^{-1})\mathfrak{s}_{\mathfrak{v}}\subseteq\mathfrak{a}\oplus\mathfrak{n}$
is orthogonal to
$\operatorname{Ad}(g)^{*}\mathfrak{v}=e^{-t\operatorname{ad}(\theta\xi)}\mathfrak{v}$
with respect to the inner product $\langle\cdot,\cdot\rangle$. Given any
$\eta\in\mathfrak{v}$, we have

$\displaystyle e^{-t\operatorname{ad}(\theta\xi)}\eta\equiv{}$
$\displaystyle\eta-t[\theta\xi,\eta]\pmod{\theta\mathfrak{n}}$
$\displaystyle={}$ $\displaystyle\eta-t\langle\xi,\eta\rangle
H_{\alpha}-\frac{t}{2}(1+\theta)[\theta\xi,\eta]\pmod{\theta\mathfrak{n}}$
$\displaystyle\equiv{}$ $\displaystyle\eta-t\langle\xi,\eta\rangle
H_{\alpha}\pmod{\mathfrak{k}_{0}\oplus\theta\mathfrak{n}}.$

Therefore, if we consider an orthonormal basis $\\{\xi,\eta\\}$ of
$\mathfrak{v}$, it is immediate that the orthogonal complement of
$\operatorname{Ad}(g^{-1})\mathfrak{s}_{\mathfrak{v}}$ in
$\mathfrak{a}\oplus\mathfrak{n}$ is
$\operatorname{span}\\{tH_{\alpha}-\xi,\eta\\}$. As a consequence,
$\tilde{\mathfrak{s}}=\operatorname{Ad}(g^{-1})\mathfrak{s}_{\mathfrak{v}}=\ker\alpha\oplus(\mathfrak{n}\ominus\mathfrak{v})\oplus\mathbb{R}(H_{\alpha}+t|\alpha|^{2}\xi)$.
The normal spaec $\mathfrak{s}^{\perp}$ is given by
$\tilde{\mathfrak{s}}^{\perp}=\mathbb{R}\eta\oplus\mathbb{R}(tH_{\alpha}-2\xi)$.

Assume $H\in\ker\alpha$. In this case, we directly have from (2) that
$\mathbb{II}(H,H)=0$.

Now, suppose that $\lambda\in\Sigma^{+}$ and
$X\in\mathfrak{g}_{\lambda}\ominus\mathfrak{v}$ is such that $1=\langle
X,X\rangle_{AN}=\frac{1}{2}|X|^{2}$. Then, $\mathbb{II}(X,X)$ satisfies
$\langle\mathbb{II}(X,X),\nu\rangle_{AN}=\frac{1}{2}\langle|X|^{2}H_{\lambda},\nu\rangle=\langle
H_{\lambda},\nu\rangle=\langle\frac{t\langle\lambda,\alpha\rangle}{2+t^{2}|\alpha|^{2}}(tH_{\alpha}-2\xi),\nu\rangle_{AN}$
for every $\nu\in\mathfrak{s}^{\perp}$, and thus
$\mathbb{II}(X,X)=\frac{t\langle\lambda,\alpha\rangle}{2+t^{2}|\alpha|^{2}}(tH_{\alpha}-2\xi)$.

Finally, consider the vector $Y=H_{\alpha}+t|\alpha|^{2}\xi$, whose norm
squared is $\langle
Y,Y\rangle_{AN}=|\alpha|^{2}+\frac{1}{2}t^{2}|\alpha|^{4}$. Note that
$(1-\theta)[\theta Y,Y]=2t|\alpha|^{4}(tH_{\alpha}-(1-\theta)\xi)$, so we
deduce from (2) that
$\mathbb{II}(Y,Y)=\frac{t|\alpha|^{4}}{2}(tH_{\alpha}-2\xi)$. As a
consequence, the normalized vector $Z=Y/|Y|_{AN}$ satisfies
$\mathbb{II}(Z,Z)=\frac{t|\alpha|^{2}}{2+t^{2}|\alpha|^{2}}(tH_{\alpha}-2\xi)$.
From these calculations, we obtain that the mean curvature of $\tilde{S}$ at
$o$ is given by

$\begin{split}\tilde{\mathcal{H}}_{t}={}&\frac{t}{2+t^{2}|\alpha|^{2}}\bigg{(}\sum_{\lambda\in\Sigma^{+}\setminus\\{\alpha\\}}(\dim\mathfrak{g}_{\lambda})\langle\lambda,\alpha\rangle+(\dim\mathfrak{g}_{\alpha}-1)|\alpha|^{2}\bigg{)}(tH_{\alpha}-2\xi)\\\
={}&\frac{t}{2+t^{2}|\alpha|^{2}}(\langle
2\delta,\alpha\rangle-|\alpha|^{2})(tH_{\alpha}-2\xi)\\\
={}&\frac{t|\alpha|^{2}}{2+t^{2}|\alpha|^{2}}(\dim\mathfrak{g}_{\alpha}+2\dim\mathfrak{g}_{2\alpha}-1)(tH_{\alpha}-2\xi).\end{split}$

Finally, note that the existence of an abelian plane inside
$\mathfrak{g}_{\alpha}$ implies that the integer
$\dim\mathfrak{g}_{\alpha}+2\dim\mathfrak{g}_{2\alpha}-1$ is positive, so the
orbit through $\operatorname{Exp}(t\xi)$ is minimal if and only if $t=0$, as
desired. ∎

In particular, the homogeneous foliation induced by $S_{\mathfrak{v}}$ is
never harmonic independently of the choice of $\mathfrak{v}$. From here,
Corollary 1.1 follows immediately.

######  Corollary 3.5.

No polar homogeneous foliation constructed as in case (i) of Theorem B is
orbit equivalent to a homogeneous foliation given in case (ii).

##  4\. Proof of Theorem B

Now we prove that the examples appearing in Theorem B are the only examples of
codimension two homogeneous polar foliations on symmetric spaces of noncompact
type.

From [2, Proposition 2.2] we obtain:

######  Proposition 4.1.

Let $M$ be a Hadamard manifold, and let $H$ be a connected closed subgroup of
the isometry group of $M$ acting on $M$ in such a way that the orbits of $H$
form a foliation. Then, all the orbits of $H$ are principal, and there is a
connected closed solvable group $S$ acting isometrically on $M$ whose orbits
coincide with the orbits of $H$.

Let $M=G/K$ be a symmetric space of noncompact type. We assume the notation
introduced in Subsection 2.1. Thus, $K$ is the isotropy group at $o\in M$, we
have a Cartan decomposition $\mathfrak{g}=\mathfrak{k}\oplus\mathfrak{p}$, a
choice of maximal abelian subspace $\mathfrak{a}$ of $\mathfrak{p}$ that
determines a root space decomposition
$\mathfrak{g}=\mathfrak{g}_{0}\oplus\bigl{(}\bigoplus_{\lambda\in\Sigma}\mathfrak{g}_{\lambda}\bigr{)}$,
and a positivity criterion that selects a set of positive roots $\Sigma^{+}$.
We denote by $\Lambda$ the set of simple roots. We define
$\mathfrak{n}=\bigoplus_{\lambda\in\Sigma^{+}}\mathfrak{g}_{\lambda}$, and
recall that $\mathfrak{k}_{0}=\mathfrak{g}_{0}\cap\mathfrak{k}$.

Assume that $H$ is a connected closed subgroup of the isometry group $G$ that
acts polarly on $M$, and that the orbits of $H$ on $M$ induce a foliation.
Proposition 4.1 says that there exists a solvable subgroup $S$ of $G$ whose
orbits coincide with the orbits of $H$. Let $\mathfrak{s}$ be the Lie algebra
of $S$. Then, $\mathfrak{s}$ is contained in a Borel subalgebra $\mathfrak{b}$
of $\mathfrak{g}$. See Subsection 2.3 for further details. The next result
states that we may assume that $\mathfrak{s}$ is contained in a maximally
noncompact Borel subalgebra.

######  Proposition 4.2.

The leaves of a homogeneous polar foliation on $M=G/K$ coincide, up to
isometric congruence, with the orbits of a connected closed solvable subgroup
$S$ of $G$ whose Lie algebra $\mathfrak{s}$ is contained in a maximally
noncompact Borel subalgebra of the form
$\mathfrak{t}\oplus\mathfrak{a}\oplus\mathfrak{n}$, where
$\mathfrak{t}\subseteq\mathfrak{k}_{0}$ is abelian.

###### Proof.

By Proposition 4.1 there is a solvable subgroup $S$ of $G$ whose orbits form
the homogeneous polar foliation under investigation. According to Proposition
2.2, we may assume that $\mathfrak{s}$ is contained in a maximal solvable
subalgebra of the form
$\tilde{\mathfrak{t}}\oplus\mathfrak{a}_{\Phi}\oplus\mathfrak{n}_{\Phi}$, with
$\tilde{\mathfrak{t}}\subseteq\mathfrak{k}_{\Phi}$ an abelian subspace, and
$\Phi\subseteq\Lambda$ a subset of simple roots. In particular, the tangent
space of $S\cdot o$ at $o$, as a subspace of $\mathfrak{p}$, is contained in
$(1-\theta)(\mathfrak{a}_{\Phi}\oplus\mathfrak{n}_{\Phi})$. As a consequence,

$(\mathfrak{a}\ominus\mathfrak{a}_{\Phi})\oplus(1-\theta)(\mathfrak{n}\ominus\mathfrak{n}_{\Phi})=\Bigl{(}\bigoplus_{\alpha\in\Phi}\mathbb{R}H_{\alpha}\Bigr{)}\oplus\Bigl{(}\bigoplus_{\lambda\in\Sigma_{\Phi}^{+}}\mathfrak{p}_{\lambda}\Bigr{)}\subseteq\mathfrak{s}_{\mathfrak{p}}^{\perp},$

where
$\mathfrak{s}_{\mathfrak{p}}^{\perp}=\\{\xi\in\mathfrak{p}:\langle\xi,\mathfrak{s}\rangle=0\\}$,
according to the definition given in Proposition 2.4. Let
$\lambda\in\Sigma_{\Phi}^{+}$ be arbitrary, and $X\in\mathfrak{g}_{\lambda}$.
Then, since $H_{\lambda}$,
$(1-\theta)X\in\mathfrak{s}_{\mathfrak{p}}^{\perp}$, and the action of $S$ is
polar, it follows from Proposition 2.4 that
$[H_{\lambda},(1-\theta)X]=(1+\theta)\lvert\lambda\rvert^{2}X$ is orthogonal
to $\mathfrak{s}$. Thus, $\mathfrak{s}$ is orthogonal to
$\bigoplus_{\lambda\in\Sigma_{\Phi}^{+}}\mathfrak{k}_{\lambda}$, and is
therefore contained in
$(\tilde{\mathfrak{t}}\cap\mathfrak{k}_{0})\oplus\mathfrak{a}_{\Phi}\oplus\mathfrak{n}_{\Phi}\subseteq(\tilde{\mathfrak{t}}\cap\mathfrak{k}_{0})\oplus\mathfrak{a}\oplus\mathfrak{n}$,
with $\mathfrak{t}=\tilde{\mathfrak{t}}\cap\mathfrak{k}_{0}$ abelian. ∎

In view of Proposition 4.2, if $S$ is a closed solvable subgroup of $G$ acting
polarly on $M$ and such that its orbits induce a foliation on $M$, we may
assume from now on that the Lie algebra $\mathfrak{s}$ of $S$ is contained in
a Borel subalgebra of the form
$\mathfrak{t}\oplus\mathfrak{a}\oplus\mathfrak{n}$, where
$\mathfrak{t}\subseteq\mathfrak{k}_{0}$ is abelian. From now on we assume that
the action of $S$ on $M$ is polar, but not hyperpolar. Recall from Proposition
2.4 that $\mathfrak{s}_{\mathfrak{p}}^{\perp}$ is a Lie triple system (but not
an abelian subspace) and
$[\mathfrak{s}_{\mathfrak{p}}^{\perp},\mathfrak{s}_{\mathfrak{p}}^{\perp}]$ is
orthogonal to $\mathfrak{s}$.

Moreover,
$[\mathfrak{s}_{\mathfrak{p}}^{\perp},\mathfrak{s}_{\mathfrak{p}}^{\perp}]\oplus\mathfrak{s}_{\mathfrak{p}}^{\perp}$
is a reductive Lie algebra and the orbit through the origin of the subgroup
$S_{\mathfrak{p}}^{\perp}$ whose Lie algebra is this one is also a symmetric
space. Since it is two-dimensional and not flat, it must be homothetic to a
real hyperbolic plane $\mathbb{R}\mathsf{H}^{2}$. Because
$\mathbb{R}\mathsf{H}^{2}$ has constant curvature, it follows from [13,
Chapter IV, Theorem 4.2] that there exists a constant $C>0$ such that
$\operatorname{ad}(\xi)^{2}(\eta)=C\eta$ for any pair of orthonormal vectors
$\xi$, $\eta\in\mathfrak{s}_{\mathfrak{p}}^{\perp}$.

Our next step is to prove that $\mathfrak{s}_{\mathfrak{p}}^{\perp}$ is
contained in $\mathfrak{a}\oplus\mathfrak{p}^{1}$. In order to do this, we
consider the vector subspace

$\tilde{\mathfrak{s}}=\mathfrak{s}\oplus(\mathfrak{n}\ominus\mathfrak{n}^{1})=\mathfrak{s}\oplus\Bigl{(}\bigoplus_{\lambda\in\Sigma^{+}\setminus\Lambda}\mathfrak{g}_{\lambda}\Bigr{)}.$

Since $\mathfrak{t}\oplus\mathfrak{a}$ normalizes all root spaces and
$[\mathfrak{n},\mathfrak{n}]\subseteq\mathfrak{n}\ominus\mathfrak{n}^{1}$, it
follows that $\tilde{\mathfrak{s}}$ is a subalgebra of
$\mathfrak{t}\oplus\mathfrak{a}\oplus\mathfrak{n}$ containing $\mathfrak{s}$.
In particular,
$\mathfrak{s}_{\mathfrak{a}\oplus\mathfrak{n}}\subseteq\tilde{\mathfrak{s}}_{\mathfrak{a}\oplus\mathfrak{n}}$,
so the codimension of $\tilde{\mathfrak{s}}_{\mathfrak{a}\oplus\mathfrak{n}}$
is less than or equal to two.

######  Lemma 4.3.

Let $\mathfrak{q}$ be a Lie subalgebra of
$\mathfrak{t}\oplus\mathfrak{a}\oplus\mathfrak{n}$ and $\lambda\in\Sigma^{+}$.
If
$\mathfrak{g}_{\lambda}\subseteq\mathfrak{q}_{\mathfrak{a}\oplus\mathfrak{n}}$
and there exists
$H\in\mathfrak{a}\cap\mathfrak{q}_{\mathfrak{a}\oplus\mathfrak{n}}$ such that
$\lambda(H)\neq 0$, then $\mathfrak{g}_{\lambda}\subseteq\mathfrak{q}$.

###### Proof.

Take
$X\in\mathfrak{g}_{\lambda}\subseteq\mathfrak{q}_{\mathfrak{a}\oplus\mathfrak{n}}$.
Then there exist vectors $T$, $T^{\prime}\in\mathfrak{t}$ such that $T+H$,
$T^{\prime}+X\in\mathfrak{q}$. In particular,
$\operatorname{ad}(T)X+\lambda(H)X=[T+H,T^{\prime}+X]\in\mathfrak{q}$. This
means that the linear map
$\operatorname{ad}(T)+\lambda(H)\operatorname{id}_{\mathfrak{g}_{\lambda}}$
preserves $\mathfrak{g}_{\lambda}$ and carries $\mathfrak{g}_{\lambda}$ to
$\mathfrak{q}$. Since $T\in\mathfrak{t}$, the linear transformation
$\operatorname{ad}(T)$ is skew-adjoint, so
$\operatorname{ad}(T)+\lambda(H)\operatorname{id}_{\mathfrak{g}_{\lambda}}$ is
a linear isomorphism and it follows that
$\mathfrak{g}_{\lambda}\subseteq\mathfrak{q}$. ∎

Now we can rule out the possibility that
$\tilde{\mathfrak{s}}_{\mathfrak{a}\oplus\mathfrak{n}}$ has codimension zero.

######  Lemma 4.4.

$\tilde{\mathfrak{s}}_{\mathfrak{a}\oplus\mathfrak{n}}\neq\mathfrak{a}\oplus\mathfrak{n}$.

###### Proof.

Assume that
$\tilde{\mathfrak{s}}_{\mathfrak{a}\oplus\mathfrak{n}}=\mathfrak{a}\oplus\mathfrak{n}$.
Both $\mathfrak{a}$ and all root spaces corresponding to positive roots are
contained in $\tilde{\mathfrak{s}}_{\mathfrak{a}\oplus\mathfrak{n}}$. By Lemma
4.3, it follows that $\mathfrak{n}\subseteq\tilde{\mathfrak{s}}$.

Let $m$ denote the maximum possible level of a root. We define
$k\in\\{0,\dots,m\\}$ to be the smallest integer for which
$\mathfrak{n}^{k+1}\oplus\dots\oplus\mathfrak{n}^{m}\subseteq\mathfrak{s}$. We
want to show that $k=0$. On the contrary, assume that $k\geq 1$. Let
$\lambda\in\Sigma^{+}$ be a root of level $k$. As $\mathfrak{n}^{1}$ generates
$\mathfrak{n}$, the root space $\mathfrak{g}_{\lambda}$ is generated by
elements of the form

$\operatorname{ad}(X_{1})\cdots\operatorname{ad}(X_{k-1})X_{k},\quad
X_{i}\in\mathfrak{n}^{1}.$

Since
$\mathfrak{n}\subseteq\tilde{\mathfrak{s}}=\mathfrak{s}+(\mathfrak{n}\ominus\mathfrak{n}^{1})$,
we can choose $Y_{1},\dots,Y_{k}\in\mathfrak{n}\ominus\mathfrak{n}^{1}$ such
that $X_{i}+Y_{i}\in\mathfrak{s}$ for each $i\in\\{1,\dots,k\\}$. Hence,
$\operatorname{ad}(X_{1}+Y_{1})\cdots\operatorname{ad}(X_{k-1}+Y_{k-1})(X_{k}+Y_{k})\in\mathfrak{s}$.
By using the fact that
$[\mathfrak{n}^{r},\mathfrak{n}^{s}]\subseteq\mathfrak{n}^{r+s}$, we have

$\operatorname{ad}(X_{1}+Y_{1})\cdots\operatorname{ad}(X_{k-1}+Y_{k-1})(X_{k}+Y_{k})\equiv\\\
\operatorname{ad}(X_{1})\cdots\operatorname{ad}(X_{k-1})X_{k}\pmod{\mathfrak{n}^{k+1}\oplus\dots\oplus\mathfrak{n}^{m}},$

so we obtain
$\operatorname{ad}(X_{1})\cdots\operatorname{ad}(X_{k-1})X_{k}\in\mathfrak{s}$.
This means that $\mathfrak{g}_{\lambda}\subseteq\mathfrak{s}$, and as a
result, $\mathfrak{n}^{k}\subseteq\mathfrak{s}$, contradicting the definition
of $k$.

Therefore, $k=0$ and $\mathfrak{n}\subseteq\mathfrak{s}$. In particular,
$\mathfrak{s}_{\mathfrak{p}}^{\perp}\subseteq\mathfrak{a}$ must be an abelian
subspace, contradicting the fact that our action is not hyperpolar. Thus, the
case
$\tilde{\mathfrak{s}}_{\mathfrak{a}\oplus\mathfrak{n}}=\mathfrak{a}\oplus\mathfrak{n}$
is not possible. ∎

Before analyzing the remaining possibilities for the codimension of
$\tilde{\mathfrak{s}}_{\mathfrak{a}\oplus\mathfrak{n}}$ we need the following
result.

######  Lemma 4.5.

Assume that $V\in\mathfrak{a}\oplus\mathfrak{n}$ is nonzero and orthogonal to
$\mathfrak{s}$. Then:

  1. (i)

If $V\in\mathfrak{a}$, then
$\mathfrak{s}_{\mathfrak{p}}^{\perp}=\mathbb{R}V\oplus(1-\theta)\mathbb{R}\eta_{\alpha}$,
where $\eta_{\alpha}\in\mathfrak{g}_{\alpha}$ is nonzero and
$\alpha\in\Lambda$.

  2. (ii)

If $V\in\mathfrak{g}_{\alpha}$ for $\alpha\in\Lambda$, then
$\mathfrak{s}_{\mathfrak{p}}^{\perp}=(1-\theta)\bigl{(}\mathbb{R}V\oplus\mathbb{R}(aH_{\alpha}+\eta_{\alpha})\bigr{)}$,
where $a\in\mathbb{R}$,
$\eta_{\alpha}\in\mathfrak{g}_{\alpha}\ominus\mathbb{R}V$, and
$[V,\eta_{\alpha}]=0$.

  3. (iii)

If $V=H_{\alpha}+\xi_{\alpha}$, where $\alpha\in\Lambda$ and
$\xi_{\alpha}\in\mathfrak{g}_{\alpha}$ is a nonzero vector, and
$g=\operatorname{Exp}(-\xi_{\alpha}/\lvert\xi_{\alpha}\rvert^{2})\in N$, then
$\operatorname{Ad}(g)\mathfrak{s}\subseteq\mathfrak{t}\oplus\mathfrak{a}\oplus\mathfrak{n}$
is orthogonal to $\xi_{\alpha}$. In particular,
$(\operatorname{Ad}(g)\mathfrak{s})_{\mathfrak{p}}^{\perp}=(1-\theta)\bigl{(}\mathbb{R}\xi_{\alpha}\oplus\mathbb{R}(aH_{\alpha}+\eta_{\alpha})\bigr{)}$,
for $a\in\mathbb{R}$ and
$\eta_{\alpha}\in\mathfrak{g}_{\alpha}\ominus\mathbb{R}\xi_{\alpha}$.
Furthermore, $[\xi_{\alpha},\eta_{\alpha}]=0$.

###### Proof.

We prove (i). Assume $V\in\mathfrak{a}$. Since
$\mathfrak{a}\subseteq\mathfrak{p}$, this means
$V\in\mathfrak{s}_{\mathfrak{p}}^{\perp}$. Choose any unit vector
$\eta=\eta_{0}+\sum_{\lambda\in\Sigma^{+}}(1-\theta)\eta_{\lambda}\in\mathfrak{s}_{\mathfrak{p}}^{\perp}$
orthogonal to $V$, where $\eta_{0}\in\mathfrak{a}$ and
$\eta_{\lambda}\in\mathfrak{g}_{\lambda}$ for each $\lambda\in\Sigma^{+}$.
Since the action of $S$ is polar nonhyperpolar, $[V,\eta]$ is a nonzero vector
orthogonal to $\mathfrak{s}$. Note that

$[V,\eta]=(1+\theta)\sum_{\lambda\in\Sigma^{+}}\lambda(V)\eta_{\lambda},$

and recalling that $\theta\mathfrak{n}$ is orthogonal to $\mathfrak{s}$, we
obtain

$(1-\theta)\sum_{\lambda\in\Sigma^{+}}\lambda(V)\eta_{\lambda}\in\mathfrak{s}_{\mathfrak{p}}^{\perp}\ominus\mathbb{R}V=\mathbb{R}\eta,$

so $\eta_{0}=0$ and $\lambda(V)=\mu(V)$ for every pair of roots $\lambda$,
$\mu\in\Sigma^{+}$ such that $\eta_{\lambda},\eta_{\mu}\neq 0$.

Suppose $\mu$, $\nu\in\Sigma^{+}$ are two different roots such that
$\eta_{\mu},\eta_{\nu}\neq 0$. Hence, $\langle
H_{\mu-\nu},V\rangle=\mu(V)-\nu(V)=0$ and $\langle H_{\mu-\nu},\eta\rangle=0$,
so $H_{\mu-\nu}\in\mathfrak{s}_{\mathfrak{a}\oplus\mathfrak{n}}$. Therefore,
we may choose a $T\in\mathfrak{t}$ such that $T+H_{\mu-\nu}\in\mathfrak{s}$.
An analogous argument shows that
$\eta_{\mu}+a\eta_{\nu}\in\mathfrak{s}_{\mathfrak{a}\oplus\mathfrak{n}}$ for
$a=-\lvert\eta_{\mu}\rvert^{2}/\lvert\eta_{\nu}\rvert^{2}<0$, so there is a
vector $T^{\prime}\in\mathfrak{t}$ such that
$T^{\prime}+\eta_{\mu}+a\eta_{\nu}\in\mathfrak{s}$. Hence,

$[T,\eta_{\mu}]+a[T,\eta_{\nu}]+\langle\mu-\nu,\mu\rangle\eta_{\mu}+a\langle\mu-\nu,\nu\rangle\eta_{\nu}=[T+H_{\mu-\nu},T^{\prime}+\eta_{\mu}+a\eta_{\nu}]\in\mathfrak{s}\cap\mathfrak{n}.$

Since $\operatorname{ad}(T)$ is skew-adjoint (because $T\in\mathfrak{k}$) we
deduce $[T,\eta_{\mu}]$,
$[T,\eta_{\nu}]\in\mathfrak{s}_{\mathfrak{a}\oplus\mathfrak{n}}$. Thus
$\langle\mu-\nu,\mu\rangle\eta_{\mu}+a\langle\mu-\nu,\nu\rangle\eta_{\nu}$ is
also in $\mathfrak{s}_{\mathfrak{a}\oplus\mathfrak{n}}$. Observe that

$\begin{vmatrix}1&a\\\
\langle\mu-\nu,\mu\rangle&a\langle\mu-\nu,\nu\rangle\end{vmatrix}=-a\lvert\mu-\nu\rvert^{2}>0,$

which implies that $\eta_{\mu}+a\eta_{\nu}$ and
$\langle\mu-\nu,\mu\rangle\eta_{\mu}+a\langle\mu-\nu,\nu\rangle\eta_{\nu}$ are
linearly independent vectors in
$\mathfrak{s}_{\mathfrak{a}\oplus\mathfrak{n}}$. Therefore, $\eta_{\mu}$,
$\eta_{\nu}\in\mathfrak{s}_{\mathfrak{a}\oplus\mathfrak{n}}$ must be
orthogonal to $\eta$, contradicting the fact that they are nonzero. We thus
obtain that only one of the $\eta_{\lambda}$ can be nonzero, that is,
$\eta=(1-\theta)\eta_{\mu}\in\mathfrak{p}_{\mu}$ for some $\mu\in\Sigma^{+}$.

We now prove that $\mu$ is simple. If $\mu=\beta+\gamma$ were a sum of
positive roots $\beta$, $\gamma\in\Sigma^{+}$, then
$\eta_{\mu}\in\mathfrak{g}_{\mu}=[\mathfrak{g}_{\beta},\mathfrak{g}_{\gamma}]$,
so we may write $\eta_{\nu}=\sum_{i=1}^{k}[X_{i},Y_{i}]$, where
$X_{i}\in\mathfrak{g}_{\beta}$ and $Y_{i}\in\mathfrak{g}_{\gamma}$ for each
$i$. Clearly,
$\mathfrak{g}_{\beta}+\mathfrak{g}_{\gamma}\subseteq\mathfrak{s}_{\mathfrak{a}\oplus\mathfrak{n}}$,
which means that for each $i$ there are vectors $T_{i}$,
$T_{i}^{\prime}\in\mathfrak{t}$ such that $T_{i}+X_{i}$,
$T_{i}^{\prime}+Y_{i}\in\mathfrak{s}$. As a consequence,

$\sum_{i=1}^{k}[T_{i},Y_{i}]+[X_{i},T_{i}^{\prime}]+[X_{i},Y_{i}]=\sum_{i=1}^{k}[T_{i}+X_{i},T_{i}^{\prime}+Y_{i}]\in\mathfrak{s}.$

Note that each $[T_{i},Y_{i}]$ is in
$\mathfrak{g}_{\gamma}\subseteq\mathfrak{s}_{\mathfrak{a}\oplus\mathfrak{n}}$
and each $[X_{i},T_{i}^{\prime}]$ is in
$\mathfrak{g}_{\beta}\subseteq\mathfrak{s}_{\mathfrak{a}\oplus\mathfrak{n}}$,
which implies that
$\eta_{\mu}=\sum_{i}[X_{i},Y_{i}]\in\mathfrak{s}_{\mathfrak{a}\oplus\mathfrak{n}}$,
contradicting that $\eta_{\mu}$ is nonzero. We deduce that $\mu\in\Lambda$, so
the first assertion is proved.

Now we prove (ii). Assume $V\in\mathfrak{g}_{\alpha}$ for a simple root
$\alpha\in\Lambda$. Choose a nonzero vector
$\eta=\eta_{0}+\sum_{\lambda\in\Sigma^{+}}(1-\theta)\eta_{\lambda}\in\mathfrak{s}_{\mathfrak{p}}^{\perp}$,
where $\eta_{0}\in\mathfrak{a}$ and $\eta_{\lambda}\in\mathfrak{g}_{\lambda}$
for each $\lambda\in\Sigma^{+}$, such that $\langle V,\eta\rangle=\langle
V,\eta_{\alpha}\rangle=0$. Let $\mu\in\Sigma^{+}$ be a positive root with
$\eta_{\mu}\neq 0$.

For now, let us suppose that we can choose $\mu$ so that $\mu\neq\alpha$.

We first prove that $\eta_{0}\in\mathbb{R}H_{\mu}$. Assume otherwise, so there
exists a vector $H\in\mathfrak{a}$ such that $\langle H,\eta_{0}\rangle=0$ and
$\langle H,H_{\mu}\rangle=\mu(H)\neq 0$. Then
$H\in\mathfrak{s}_{\mathfrak{a}\oplus\mathfrak{n}}$, so there exists
$T\in\mathfrak{t}$ such that $T+H\in\mathfrak{s}$. On the other hand,
$\eta_{0}+a\eta_{\mu}$ is orthogonal to both $V$ and $\eta$ for
$a=-\lvert\eta_{0}\rvert^{2}/\lvert\eta_{\mu}\rvert^{2}<0$, which implies that
$\eta_{0}+a\eta_{\mu}\in\mathfrak{s}_{\mathfrak{a}\oplus\mathfrak{n}}$, and
there exists $T^{\prime}\in\mathfrak{t}$ such that
$T^{\prime}+\eta_{0}+a\eta_{\mu}\in\mathfrak{s}$. In particular,
$a[T,\eta_{\mu}]+a\mu(H)\eta_{\mu}=[T+H,T^{\prime}+\eta_{0}+a\eta_{\mu}]\in\mathfrak{s}$,
so $0=\langle
a[T,\eta_{\mu}]+a\mu(H)\eta_{\mu},\eta\rangle=a\mu(H)\lvert\eta_{\mu}\rvert^{2}$,
a contradiction.

We now prove that $\eta_{\lambda}=0$ for every
$\lambda\in\Sigma^{+}\setminus\\{\mu\\}$. If $\lambda$ is a positive root
linearly independent with $\mu$, and $\eta_{\lambda}\neq 0$, we may find
$H\in\mathfrak{a}$ such that $\mu(H)=0\neq\lambda(H)$. Since
$\eta_{0}\in\mathbb{R}H_{\mu}$, this implies that
$H\in\mathfrak{s}_{\mathfrak{a}\oplus\mathfrak{n}}$, and hence, there exists
$T\in\mathfrak{t}$ such that $T+H\in\mathfrak{s}$. Furthermore,
$\eta_{\mu}+b\eta_{\lambda}\in\mathfrak{s}_{\mathfrak{a}\oplus\mathfrak{n}}$
with $b=-\lvert\eta_{\mu}\rvert^{2}/\lvert\eta_{\lambda}\rvert^{2}<0$, and
there exists $T^{\prime}\in\mathfrak{t}$ satisfying
$T^{\prime}+\eta_{\mu}+b\eta_{\lambda}\in\mathfrak{s}$. As a consequence,
$[T,\eta_{\mu}]+b[T,\eta_{\lambda}]+b\lambda(H)\eta_{\lambda}=[T+H,T^{\prime}+\eta_{\mu}+b\eta_{\lambda}]\in\mathfrak{s}$.
In particular,
$0=\langle[T,\eta_{\mu}]+b[T,\eta_{\lambda}]+b\lambda(H)\eta_{\lambda},\eta\rangle=b\lambda(H)\lvert\eta_{\lambda}\rvert^{2}$,
a contradiction. On the other hand, if $2\mu\in\Sigma^{+}$ and
$\eta_{2\mu}\neq 0$, a similar argument yields that there is
$T\in\mathfrak{t}$ such that $T+H_{\mu}+a\eta_{\mu}\in\mathfrak{s}$, for
$a=-\mu(\eta_{0})/\lvert\eta_{\mu}\rvert^{2}$, and $T^{\prime}\in\mathfrak{t}$
such that $T^{\prime}+\eta_{\mu}+b\eta_{2\mu}\in\mathfrak{s}$, where
$b=-\lvert\eta_{\mu}\rvert^{2}/\lvert\eta_{2\mu}\rvert^{2}<0$. Thus,

$\displaystyle 0$
$\displaystyle{}=\langle[T+H_{\mu}+a\eta_{\mu},T^{\prime}+\eta_{\mu}+b\eta_{2\mu}],\eta\rangle$
$\displaystyle{}=\langle[T,\eta_{\mu}]+b[T,\eta_{2\mu}]+\lvert\mu\rvert^{2}\eta_{\mu}+2b\lvert\mu\rvert^{2}\eta_{2\mu}+a[\eta_{\mu},T^{\prime}],\eta\rangle=-\lvert\mu\rvert^{2}\lvert\eta_{\mu}\rvert^{2},$

contradicting our choice of $\mu$. This implies $\eta_{2\mu}=0$ (and an
analogous argument shows that $\eta_{\mu/2}=0$ in the case that
$\mu/2\in\Sigma^{+}$).

To summarize, we have obtained $\eta=aH_{\mu}+(1-\theta)\eta_{\mu}$ for some
constant $a\in\mathbb{R}$, and $\mu$ is a root different from $\alpha$ with
$\eta_{\mu}\neq 0$.

Assume $a\neq 0$. Since $\mathfrak{s}_{\mathfrak{p}}^{\perp}$ is a Lie triple
system, $\operatorname{ad}((1-\theta)V)^{2}\eta$ is proportional to $\eta$.
Since for any $H\in\mathfrak{a}$ we have

$\displaystyle\langle\operatorname{ad}((1-\theta)V)^{2}\eta,H\rangle$
$\displaystyle{}=-\langle[(1-\theta)V,\eta],[H,(1-\theta)V]\rangle$
$\displaystyle{}=-\langle(1+\theta)\bigl{(}-a\langle\alpha,\mu\rangle
V+[V,\eta_{\mu}]-[\theta V,\eta_{\mu}]\bigr{)},\alpha(H)(1+\theta)V\rangle$
$\displaystyle{}=2a\langle\alpha,\mu\rangle\lvert V\rvert^{2}\alpha(H)=\langle
2a\langle\alpha,\mu\rangle\lvert V\rvert^{2}H_{\alpha},H\rangle,$

it follows that $\mu=\alpha$ or $\mu=2\alpha$. This last case is not possible,
because by Proposition 2.4, $(1+\theta)(-2a\lvert\alpha\rvert^{2}V-[\theta
V,\eta_{2\alpha}])=[(1-\theta)V,aH_{2\alpha}+(1-\theta)\eta_{2\alpha}]$ would
be orthogonal to $\mathfrak{s}$. This would imply that $[\theta
V,\eta_{2\alpha}]$ is proportional to $V$, and thus,

$0=[V,[\theta V,\eta_{2\alpha}]]=-[\eta_{2\alpha},[V,\theta V]]=2\lvert
V\rvert^{2}\lvert\alpha\rvert^{2}\eta_{2\alpha},$

contradicting the fact that $\eta_{2\alpha}\neq 0$. We conclude that
$\mu=\alpha$.

Suppose now that $a=0$, so $\eta\in\mathfrak{p}_{\mu}$. Since
$\mathfrak{a},\mathfrak{g}_{\alpha+\mu}\subseteq\mathfrak{s}_{\mathfrak{a}\oplus\mathfrak{n}}$,
we obtain $\mathfrak{g}_{\alpha+\mu}\subseteq\mathfrak{s}$ by Lemma 4.3. The
vector $[(1-\theta)V,(1-\theta)\eta_{\mu}]=(1+\theta)([V,\eta_{\mu}]-[\theta
V,\eta_{\mu}])$ is nonzero and orthogonal to $\mathfrak{s}$. Combining this
with the fact that $\mathfrak{g}_{\alpha+\mu}\subseteq\mathfrak{s}$, we get
$[V,\eta_{\mu}]=0$, so $(1+\theta)[\theta V,\eta_{\mu}]$ is nonzero and
orthogonal to $\mathfrak{s}$. This now implies $\mu-\alpha\in\Sigma^{+}$ and
$[\theta V,\eta_{\mu}]\in(\mathfrak{a}\oplus\mathfrak{n})\ominus\mathfrak{s}$.
Furthermore, we must have $\mu=2\alpha$, and thus, $[\theta
V,\eta_{\mu}]\in((\mathfrak{a}\oplus\mathfrak{n})\ominus\mathfrak{s})\cap\mathfrak{g}_{\alpha}=\mathbb{R}V$.
Therefore,

$0=[V,[\theta V,\eta_{2\alpha}]]=-[\theta
V,[\eta_{2\alpha},V]]-[\eta_{2\alpha},[V,\theta V]]=-[\lvert
V\rvert^{2}H_{\alpha},\eta_{2\alpha}]=-2\lvert\alpha\rvert^{2}\lvert
V\rvert^{2}\eta_{2\alpha},$

which yields a contradiction.

We now assume $\eta_{\mu}=0$ for every
$\mu\in\Sigma^{+}\setminus\\{\alpha\\}$. As a consequence,
$\eta=\eta_{0}+(1-\theta)\eta_{\alpha}$, with $\langle
V,\eta_{\alpha}\rangle=0$. We only need to prove that
$\eta_{0}\in\mathbb{R}H_{\alpha}$. Indeed, if $\eta_{0}$ is not proportional
to $H_{\alpha}$, there exists $H\in\mathfrak{a}$ such that $\langle
H,\eta_{0}\rangle=0$ and $\alpha(H)\neq 0$. Therefore,
$H\in\mathfrak{s}_{\mathfrak{a}\oplus\mathfrak{n}}$, so we may find
$T\in\mathfrak{t}$ satisfying $T+H\in\mathfrak{s}$. Similarly, by taking
$\eta_{0}+x\eta_{\alpha}$ with
$x=-\lvert\eta_{0}\rvert^{2}/\lvert\eta_{\alpha}\rvert^{2}<0$, we obtain
$\eta_{0}+x\eta_{\alpha}\in\mathfrak{s}_{\mathfrak{a}\oplus\mathfrak{n}}$, so
$T^{\prime}+\eta_{0}+x\eta_{\alpha}\in\mathfrak{s}$ for an adequate
$T^{\prime}\in\mathfrak{t}$. Thus,
$0=\langle[T+H,T^{\prime}+\eta_{0}+x\eta_{\alpha}],\eta\rangle=x\alpha(H)\lvert\eta_{\alpha}\rvert^{2}$,
contradiction. Hence, we may write $\eta=aH_{\alpha}+(1-\theta)\eta_{\alpha}$
with $a\in\mathbb{R}$,
$\eta_{\alpha}\in\mathfrak{g}_{\alpha}\ominus\mathbb{R}V$, and
$\mathfrak{s}_{\mathfrak{p}}^{\perp}=\mathbb{R}(1-\theta)V\oplus\mathbb{R}\eta$.

If $a=0$, then note that $\mathfrak{a}$,
$\mathfrak{g}_{2\alpha}\subseteq\mathfrak{s}_{\mathfrak{a}\oplus\mathfrak{n}}$,
and Lemma 4.3 implies $\mathfrak{g}_{2\alpha}\subseteq\mathfrak{s}$. Together
with the fact that
$[(1-\theta)V,(1-\theta)\eta_{\alpha}]=(1+\theta)([V,\eta_{\alpha}]-[\theta
V,\eta_{\alpha}])$ is orthogonal to $\mathfrak{s}$ by Proposition 2.4, we get
$[V,\eta_{\alpha}]=0$.

If $a\neq 0$, we can take the triple bracket $[\eta,[\eta,(1-\theta)V]]$,
which is in
$\mathfrak{s}_{\mathfrak{p}}^{\perp}\subseteq\mathfrak{a}\oplus\mathfrak{p}^{1}$.
Then, for any $X\in\mathfrak{g}_{2\alpha}$:

$\displaystyle 0$
$\displaystyle{}=\langle[\eta,[\eta,(1-\theta)V]],X\rangle=\langle[\eta,(1-\theta)V],[\eta,X]]\rangle$
$\displaystyle{}=\langle(1+\theta)(a\lvert\alpha\rvert^{2}V+[\eta_{\alpha},V]-[\theta\eta_{\alpha},V]),2a\lvert\alpha\rvert^{2}X-[\theta\eta_{\alpha},X]\rangle$
$\displaystyle{}=-a\lvert\alpha\rvert^{2}\langle
V,[\theta\eta_{\alpha},X]\rangle+2a\lvert\alpha\rvert^{2}\langle[\eta_{\alpha},V],X\rangle=-3a\lvert\alpha\rvert^{2}\langle[V,\eta_{\alpha}],X\rangle,$

and this yields $[V,\eta_{\alpha}]=0$, as stated. This finishes the proof of
(ii).

To prove (iii), assume $V=H_{\alpha}+\xi_{\alpha}$ for a nonzero
$\xi_{\alpha}\in\mathfrak{g}_{\alpha}$, where $\alpha\in\Lambda$, and consider
$g=\operatorname{Exp}(-\xi_{\alpha}/\lvert\xi_{\alpha}\rvert^{2})\in N$. Then,
the isomorphism $\operatorname{Ad}(g)$ preserves the subalgebra
$\mathfrak{t}\oplus\mathfrak{a}\oplus\mathfrak{n}$, and therefore,
$\operatorname{Ad}(g)\mathfrak{s}\subseteq\mathfrak{t}\oplus\mathfrak{a}\oplus\mathfrak{n}$
is the Lie algebra of $gSg^{-1}$, which induces a homogeneous polar foliation
on $M$ with a non flat section. Observe that

$\operatorname{Ad}(g^{-1})^{*}(H_{\alpha}+\xi_{\alpha})=e^{-\frac{1}{\lvert\xi_{\alpha}\rvert^{2}}\operatorname{ad}(\theta\xi_{\alpha})}\xi_{\alpha}=\xi_{\alpha}-\frac{\lvert\alpha\rvert^{2}}{2\lvert\xi_{\alpha}\rvert^{2}}\theta\xi_{\alpha},$

which means that
$\operatorname{Ad}(g)\mathfrak{s}\subseteq\mathfrak{t}\oplus\mathfrak{a}\oplus\mathfrak{n}$
is orthogonal to $\xi_{\alpha}$, as desired. The rest of the assertion follows
from (ii). ∎

Now we continue with the proof of Theorem B. Recall that
$\tilde{\mathfrak{s}}=\mathfrak{s}+(\mathfrak{n}\ominus\mathfrak{n}^{1})$.
According to Lemma 4.4,
$\tilde{\mathfrak{s}}_{\mathfrak{a}\oplus\mathfrak{n}}\neq\mathfrak{a}\oplus\mathfrak{n}$,
so either $\tilde{\mathfrak{s}}_{\mathfrak{a}\oplus\mathfrak{n}}$ has
codimension one or two in $\mathfrak{a}\oplus\mathfrak{n}$.

If $\tilde{\mathfrak{s}}_{\mathfrak{a}\oplus\mathfrak{n}}$ has codimension
two, we have
$\tilde{\mathfrak{s}}_{\mathfrak{a}\oplus\mathfrak{n}}=\mathfrak{s}_{\mathfrak{a}\oplus\mathfrak{n}}$,
which is equivalent to
$\mathfrak{s}_{\mathfrak{p}}^{\perp}\subseteq\mathfrak{a}\oplus\mathfrak{p}^{1}$.

On the other hand, if $\tilde{\mathfrak{s}}_{\mathfrak{a}\oplus\mathfrak{n}}$
is a codimension one subspace of $\mathfrak{a}\oplus\mathfrak{n}$, then by [3,
Proposition 5.4] we have
$\tilde{\mathfrak{s}}_{\mathfrak{a}\oplus\mathfrak{n}}=(\mathfrak{a}\oplus\mathfrak{n})\ominus\mathbb{R}\xi$,
where $\xi$ satisfies one of the following possibilities:111Note that Berndt
and Tamaru’s proof does not rely on the additional condition that they impose
on $\mathfrak{s}$, namely, that $\mathfrak{s}\cap\mathfrak{t}=0$.

  1. (i)

$\xi\in\mathfrak{a}$.

  2. (ii)

$\xi\in\mathfrak{g}_{\alpha}$ for a simple root $\alpha\in\Lambda$.

  3. (iii)

$\xi=H_{\alpha}+\xi_{\alpha}$, where $\xi_{\alpha}\in\mathfrak{g}_{\alpha}$ is
a nonzero vector and $\alpha\in\Lambda$.

Note that $\xi$ is also orthogonal to $\mathfrak{s}$. Hence, by Lemma 4.5, we
obtain that $\mathfrak{s}_{\mathfrak{p}}^{\perp}$ may be assumed to be in
$\mathfrak{a}\oplus\mathfrak{p}^{1}$ after conjugation by an element of $N$.
The next step is to determine the orthogonal projection
$\mathfrak{s}_{\mathfrak{a}\oplus\mathfrak{n}}$.

######  Lemma 4.6.

The action of $S$ is orbit equivalent to the action of a connected closed
subgroup $\bar{S}$ whose Lie algebra $\bar{\mathfrak{s}}$ is contained in
$\mathfrak{t}\oplus\mathfrak{a}\oplus\mathfrak{n}$, the normal space
$\bar{\mathfrak{s}}_{\mathfrak{p}}^{\perp}$ is contained in
$\mathfrak{a}\oplus\mathfrak{p}^{1}$, and
$\bar{\mathfrak{s}}_{\mathfrak{p}}^{\perp}\cap\mathfrak{p}^{1}\neq 0$.
Equivalently, the orthogonal projection of
$\bar{\mathfrak{s}}_{\mathfrak{p}}^{\perp}$ on $\mathfrak{a}$ is not two-
dimensional.

###### Proof.

Assume $\mathfrak{s}_{\mathfrak{p}}^{\perp}\cap\mathfrak{p}^{1}$ is trivial.
Let
$\Psi=\\{\alpha\in\Lambda\colon\pi_{\mathfrak{g}_{\alpha}}(\mathfrak{s}_{\mathfrak{p}}^{\perp})\neq
0\\}$, where
$\pi_{\mathfrak{g}_{\alpha}}\colon\mathfrak{g}\to\mathfrak{g}_{\alpha}$
denotes the orthogonal projection. Since the action is not hyperpolar, $\Psi$
is a nonempty subset of $\Lambda$. We prove
$\mathfrak{a}^{\Psi}=\bigoplus_{\alpha\in\Psi}\mathbb{R}H_{\alpha}\subseteq\pi_{\mathfrak{a}}(\mathfrak{s}_{\mathfrak{p}}^{\perp})$.
Here $\pi_{\mathfrak{a}}$ denotes the orthogonal projection onto
$\mathfrak{a}$. Indeed, assume
$H\in\mathfrak{a}\ominus\pi_{\mathfrak{a}}(\mathfrak{s}_{\mathfrak{p}}^{\perp})$.
Then there exists a vector $T\in\mathfrak{t}$ such that $T+H\in\mathfrak{s}$.
On the other hand, let $\alpha\in\Psi$. We may find two vectors
$\xi=\xi_{0}+\sum_{\beta\in\Lambda}(1-\theta)\xi_{\beta}$ and
$\eta=\eta_{0}+\sum_{\beta\in\Lambda}(1-\theta)\eta_{\beta}$ in
$\mathfrak{s}_{\mathfrak{p}}^{\perp}$ such that $\xi_{\alpha}\neq 0$ and
$\langle\xi_{\alpha},\eta_{\alpha}\rangle=0$. By our assumption, $\xi_{0}$ and
$\eta_{0}$ are linearly independent, which implies that there exist unique
constants $x$, $y\in\mathbb{R}$ such that
$\xi_{\alpha}+x\xi_{0}+y\eta_{0}\in\mathfrak{s}_{\mathfrak{a}\oplus\mathfrak{n}}$.
In particular, we may find $T^{\prime}\in\mathfrak{t}$ satisfying
$T^{\prime}+\xi_{\alpha}+x\xi_{0}+y\eta_{0}\in\mathfrak{s}$. Thus,
$[T,\xi_{\alpha}]+\alpha(H)\xi_{\alpha}=[T+H,T^{\prime}+\xi_{\alpha}+x\xi_{0}+y\eta_{0}]\in\mathfrak{s}$.
Taking the inner product with $\xi$, we deduce $\alpha(H)=0$. All in all, we
obtain
$\mathfrak{a}\ominus\pi_{\mathfrak{a}}(\mathfrak{s}_{\mathfrak{p}}^{\perp})\subseteq\mathfrak{a}_{\Psi}=\cap_{\alpha\in\Psi}\ker\alpha$,
so $\mathfrak{a}^{\Psi}\subseteq\pi_{\mathfrak{a}}(\mathfrak{s}^{\perp})$ (in
particular, $\Psi$ has either one or two elements).

In order to prove the result, we assume first that for all $\alpha\in\Psi$,
the orthogonal projection
$\pi_{\mathfrak{g}_{\alpha}}(\mathfrak{s}_{\mathfrak{p}}^{\perp})$ is one-
dimensional. Thus, we can take two orthogonal vectors
$\xi=\xi_{0}+\sum_{\alpha\in\Psi}(1-\theta)\xi_{\alpha}$ and
$\eta=\eta_{0}+\sum_{\alpha\in\Psi}(1-\theta)\eta_{\alpha}$ that span
$\mathfrak{s}_{\mathfrak{p}}^{\perp}$. Since the action is polar
nonhyperpolar, the vector $[\xi,\eta]$ is nonzero and orthogonal to
$\mathfrak{s}$. Observe that, because $\Psi\subseteq\Lambda$,
$[\theta\xi_{\alpha},\eta_{\alpha}]\in\mathfrak{a}$ for all $\alpha\in\Psi$,
and $\alpha-\beta\notin\Sigma$, we have

$[\xi,\eta]=(1+\theta)\left(\sum_{\alpha\in\Psi}\Bigl{(}\alpha(\xi_{0})\eta_{\alpha}-\alpha(\eta_{0})\xi_{\alpha}\Bigr{)}+\sum_{\alpha,\beta\in\Psi}[\xi_{\alpha},\eta_{\beta}]\right).$

Since $\theta\mathfrak{n}$ and $\mathfrak{s}$ are orthogonal, we obtain

$(1-\theta)\left(\sum_{\alpha\in\Psi}\Big{(}\alpha(\xi_{0})\eta_{\alpha}-\alpha(\eta_{0})\xi_{\alpha}\Bigr{)}+\sum_{\alpha,\beta\in\Psi}[\xi_{\alpha},\eta_{\beta}]\right)\in\mathfrak{s}_{\mathfrak{p}}^{\perp}\subseteq\mathfrak{a}\oplus\mathfrak{p}^{1},$

which means that all terms in $\mathfrak{p}^{2}$ cancel out and
$\mathfrak{s}_{\mathfrak{p}}^{\perp}\cap\mathfrak{p}^{1}\neq 0$, a
contradiction.

Now, assume that there exists $\alpha\in\Psi$ such that
$\pi_{\mathfrak{g}_{\alpha}}(\mathfrak{s}_{\mathfrak{p}}^{\perp})$ is two-
dimensional. Since
$H_{\alpha}\in\pi_{\mathfrak{a}}(\mathfrak{s}_{\mathfrak{p}}^{\perp})$, we may
find
$H_{\alpha}+\sum_{\beta\in\Psi}(1-\theta)\xi_{\beta}\in\mathfrak{s}_{\mathfrak{p}}^{\perp}$,
with each $\xi_{\beta}\in\mathfrak{g}_{\beta}$, and $\xi_{\alpha}\neq 0$, for
dimension reasons. Consider the element
$g=\operatorname{Exp}(-\xi_{\alpha}/\lvert\xi_{\alpha}\rvert^{2})\in N$. Then
the action of $S$ is orbit equivalent to the action of $gSg^{-1}$, whose Lie
algebra is
$\operatorname{Ad}(g)\mathfrak{s}\subseteq\mathfrak{t}\oplus\mathfrak{a}\oplus\mathfrak{n}$.
Note that the equality

$\operatorname{Ad}(g^{-1})^{*}\biggl{(}H_{\alpha}+\sum_{\beta_{\in}\Psi}\xi_{\beta}\biggr{)}=\sum_{\beta\in\Psi}\xi_{\beta}-\frac{\lvert\alpha\rvert^{2}}{2\lvert\xi_{\alpha}\rvert^{2}}\theta\xi_{\alpha}$

and $\langle\theta\mathfrak{n},\mathfrak{s}\rangle=0$ imply
$\sum_{\beta\in\Psi}(1-\theta)\xi_{\beta}\in(\operatorname{Ad}(g)\mathfrak{s})_{\mathfrak{p}}^{\perp}\cap\mathfrak{p}^{1}$.

To conclude, it suffices to prove that
$(\operatorname{Ad}(g)\mathfrak{s})_{\mathfrak{p}}^{\perp}\subseteq\mathfrak{a}\oplus\mathfrak{p}^{1}$.
This is the case if the projection of
$\operatorname{Ad}(g)\mathfrak{s}+(\mathfrak{n}\ominus\mathfrak{n}^{1})$ onto
$\mathfrak{a}\oplus\mathfrak{n}$ has codimension 2. As the projection of
$\operatorname{Ad}(g)\mathfrak{s}+(\mathfrak{n}\ominus\mathfrak{n}^{1})$ onto
$\mathfrak{a}\oplus\mathfrak{n}$ cannot be $\mathfrak{a}\oplus\mathfrak{n}$ by
Lemma 4.4, our assertion is false whenever this projection is of codimension
one, that is, when the orthogonal complement of
$\operatorname{Ad}(g)\mathfrak{s}$ in $\mathfrak{a}\oplus\mathfrak{n}$ is
spanned by $\sum_{\beta\in\Psi}\xi_{\beta}$. By [3, Proposition 5.4],
$\xi_{\beta}=0$ for all simple roots $\beta\neq\alpha$, and by Lemma 4.5(ii)
we have
$(\operatorname{Ad}(g)\mathfrak{s})_{\mathfrak{p}}^{\perp}=(1-\theta)\bigl{(}\mathbb{R}\xi_{\alpha}\oplus(aH_{\alpha}+\eta_{\alpha})\bigr{)}$
for $a\in\mathbb{R}$ and $\eta_{\alpha}\in\mathfrak{g}_{\alpha}$. Thus,
$(\operatorname{Ad}(g)\mathfrak{s})_{\mathfrak{p}}^{\perp}\subseteq\mathfrak{a}\oplus\mathfrak{p}^{1}$,
contradicting the fact that the projection of
$\operatorname{Ad}(g)\mathfrak{s}+(\mathfrak{n}\ominus\mathfrak{n}^{1})$ onto
$\mathfrak{a}\oplus\mathfrak{n}$ has codimension one. ∎

Due to the previous lemma, we may assume that
$\mathfrak{s}_{\mathfrak{p}}^{\perp}\cap\mathfrak{p}^{1}$ is a nonzero
subspace of $\mathfrak{g}$.

First we need:

######  Lemma 4.7.

Let $S$ be a closed subgroup of $G$ whose Lie algebra $\mathfrak{s}$ satisfies
$\mathfrak{s}\subseteq\mathfrak{t}\oplus\mathfrak{a}\oplus\mathfrak{n}$, and
the orbits of $S$ form a homogeneous foliation on $M$. Let $V\in\mathfrak{n}$
be a vector such that $(1-\theta)V\in\mathfrak{s}_{\mathfrak{p}}^{\perp}$ and
$g=\operatorname{Exp}(V)\in N$. Then
$\mathfrak{s}\cap\mathfrak{t}=\operatorname{Ad}(g)(\mathfrak{s}\cap\mathfrak{t})=\operatorname{Ad}(g)(\mathfrak{s})\cap\mathfrak{k}$.

###### Proof.

Since the orbit $S\cdot o$ is principal, we have
$[\mathfrak{s}\cap\mathfrak{t},\mathfrak{s}_{\mathfrak{p}}^{\perp}]=0$. Hence,
$[\mathfrak{s}\cap\mathfrak{t},(1-\theta)V]=(1-\theta)[\mathfrak{s}\cap\mathfrak{t},V]=0$,
which means $\operatorname{ad}(V)(\mathfrak{s}\cap\mathfrak{t})=0$ and
$\operatorname{Ad}(g)(\mathfrak{s}\cap\mathfrak{t})=e^{\operatorname{ad}(V)}(\mathfrak{s}\cap\mathfrak{t})=\mathfrak{s}\cap\mathfrak{t}$.
On the other hand, $\mathfrak{s}\cap\operatorname{Ad}(g^{-1})\mathfrak{k}$ is
the isotropy algebra of $S$ at $g^{-1}\cdot o$, and since all orbits have the
same type, it follows that
$\dim\operatorname{Ad}(g)\mathfrak{s}\cap\mathfrak{k}=\dim\mathfrak{s}\cap\operatorname{Ad}(g^{-1})\mathfrak{k}=\dim\mathfrak{s}\cap\mathfrak{k}=\dim\mathfrak{s}\cap\mathfrak{t}$.
Since
$\mathfrak{s}\cap\mathfrak{t}=\operatorname{Ad}(g)(\mathfrak{s}\cap\mathfrak{t})\subseteq\operatorname{Ad}(g)(\mathfrak{s})\cap\mathfrak{k}$,
the equality follows. ∎

The next result is needed later to handle the two examples in Theorem B
simultaneously.

######  Proposition 4.8.

Let $S$ be a connected closed subgroup of $G$ inducing a homogeneous polar
foliation on $M$. Assume that its Lie algebra is contained in a maximally
noncompact Borel subalgebra
$\mathfrak{t}\oplus\mathfrak{a}\oplus\mathfrak{n}$, and
$\mathfrak{s}_{\mathfrak{a}\oplus\mathfrak{n}}=\mathfrak{z}\oplus(\mathfrak{n}\ominus\mathfrak{v}_{\alpha})$,
where $\mathfrak{v}_{\alpha}$ is an abelian subspace of
$\mathfrak{g}_{\alpha}$, $\alpha\in\Lambda$, and $\mathfrak{z}$ is a subspace
of $\mathfrak{a}$. Let $\tilde{S}$ be the connected Lie subgroup of $G$ whose
Lie algebra is
$\tilde{\mathfrak{s}}=\mathfrak{z}\oplus(\mathfrak{n}\ominus\mathfrak{v}_{\alpha})$.
Then, $S$ and $\tilde{S}$ have the same orbits.

###### Proof.

Denote by $\mathfrak{s}_{\mathfrak{t}}$ the orthogonal projection of
$\mathfrak{s}$ onto $\mathfrak{t}$. We start by proving that
$\hat{\mathfrak{s}}=\mathfrak{s}_{\mathfrak{t}}\oplus\tilde{\mathfrak{s}}$ is
a Lie subalgebra of $\mathfrak{g}$. Since
$[\mathfrak{g}_{\lambda},\mathfrak{g}_{\mu}]\subseteq\mathfrak{g}_{\lambda+\mu}$,
for $\lambda$, $\mu\in\Sigma^{+}$, and $\mathfrak{t}$ is abelian, centralizes
$\mathfrak{a}$, and normalizes each root space, in reality this amounts to
proving
$[\mathfrak{s}_{\mathfrak{t}},\mathfrak{g}_{\alpha}\ominus\mathfrak{v}_{\alpha}]\subseteq\mathfrak{g}_{\alpha}\ominus\mathfrak{v}_{\alpha}$.

Let $U$, $V\in\mathfrak{v}_{\alpha}$ and $T\in\mathfrak{s}_{\mathfrak{t}}$. We
choose $X\in{\mathfrak{a}\oplus\mathfrak{n}}$, such that $T+X\in\mathfrak{s}$.
Since the action of $S$ is polar, we know that
$[\mathfrak{s}_{\mathfrak{p}}^{\perp},\mathfrak{s}_{\mathfrak{p}}^{\perp}]$ is
perpendicular to $\mathfrak{s}$. Thus,

$\displaystyle 0$
$\displaystyle{}=\langle[(1-\theta)U,(1-\theta)V],\,T+X\rangle=\langle-(1+\theta)[\theta
U,V],\,T\rangle$ $\displaystyle{}=-2\langle[U,\theta V],\,T\rangle=-2\langle
U,[V,T]\rangle.$

This proves
$[\mathfrak{s}_{\mathfrak{t}},\mathfrak{v}_{\alpha}]\subseteq\mathfrak{g_{\alpha}\ominus\mathfrak{v}_{\alpha}}$.

Let $T\in\mathfrak{s}\cap\mathfrak{t}$ and $V\in\mathfrak{v}_{\alpha}$. Since
$\mathfrak{s}\cap\mathfrak{t}\subseteq\mathfrak{s}_{\mathfrak{t}}$, we have
$[T,V]\in\mathfrak{g}_{\alpha}\ominus\mathfrak{v}_{\alpha}$. Let
$X\in\mathfrak{g}_{\alpha}\ominus\mathfrak{v}_{\alpha}$. Then there exists
$T_{X}\in\mathfrak{t}$ such that $T_{X}+X\in\mathfrak{s}$. Thus,
$[T,X]=[T,\,T_{X}+X]\in\mathfrak{s}$, and hence, $\langle[T,X],V\rangle=0$. We
have proved $[\mathfrak{s}\cap\mathfrak{t},\mathfrak{v}_{\alpha}]=0$.

Let
$T\colon\tilde{\mathfrak{s}}\to\mathfrak{s}_{\mathfrak{t}}\ominus(\mathfrak{s}\cap\mathfrak{t})$,
$X\mapsto T_{X}$, be defined by $T_{X}+X\in\mathfrak{s}$. This map is well-
defined: if $T_{X}$, $T_{X}^{\prime}\in\mathfrak{t}$ are such that $T_{X}+X$,
$T_{X}^{\prime}+X\in\mathfrak{s}$, subtracting,
$T_{X}-T_{X}^{\prime}\in\mathfrak{s}\cap\mathfrak{t}$. Note that $T$ is
surjective.

Given a nonzero $V\in\mathfrak{v}_{\alpha}$, we define
$\Phi_{V}\colon\mathfrak{g}_{\alpha}\ominus\mathfrak{v}_{\alpha}\to\mathfrak{g}_{\alpha}\ominus\mathfrak{v}_{\alpha}$
by $\Phi_{V}(X)=[T_{X},V]$. We prove that $\Phi_{V}$ is self-adjoint. Indeed,
given $X$,
$Y\in\mathfrak{g}_{\alpha}\ominus\mathfrak{v}_{\alpha}\subseteq\tilde{\mathfrak{s}}$,
we obtain $[T_{X},Y]+[X,T_{Y}]+[X,Y]=[T_{X}+X,T_{Y}+Y]\in\mathfrak{s}$, which
means

$0=\langle
V,[T_{X}+X,T_{Y}+Y]\rangle=-\langle[T_{X},V],Y\rangle+\langle[T_{Y},V],X\rangle=-\langle\Phi_{V}(X),Y\rangle+\langle
X,\Phi_{V}(Y)\rangle.$

We now prove that $\Phi_{V}=0$. Assume this is not the case, so by the
spectral theorem, there exists a nonzero vector
$X\in\mathfrak{g}_{\alpha}\ominus\mathfrak{v}_{\alpha}$ and a nonzero constant
$\lambda\in\mathbb{R}$ such that $\Phi_{V}(X)=\lambda X$.

Observe that $[V,T_{[V,X]}]=0$. Indeed,
$[V,T_{[V,X]}]\in\mathfrak{g}_{\alpha}\ominus\mathfrak{v}_{\alpha}$, and given
any $Y\in\mathfrak{g}_{\alpha}\ominus\mathfrak{v}_{\alpha}$ we obtain

$\displaystyle 0$
$\displaystyle{}=\langle[T_{Y}+Y,T_{[V,X]}+[V,X]],\,V\rangle=\langle[T_{Y},[V,X]]+[Y,T_{[V,X]}],\,V\rangle$
$\displaystyle{}=\langle[Y,T_{[V,X]}],\,V\rangle=-\langle
Y,\,[V,T_{[V,X]}]\rangle,$

which implies $[V,T_{[V,X]}]=0$.

Now, consider $g=\operatorname{Exp}(\frac{1}{\lambda}V)$ and
$Z=T_{X}+X-\frac{1}{2\lambda}\bigl{(}T_{[V,X]}+[V,X]\bigr{)}\in\mathfrak{s}$.
Then

$\displaystyle\operatorname{Ad}(g)Z$
$\displaystyle{}=e^{\frac{1}{\lambda}\operatorname{ad}(V)}\Bigl{(}T_{X}+X-\frac{1}{2\lambda}T_{[V,X]}-\frac{1}{2\lambda}[V,X]\Bigr{)}$
$\displaystyle{}=T_{X}+X-\frac{1}{2\lambda}T_{[V,X]}-\frac{1}{2\lambda}[V,X]+\dfrac{1}{\lambda}\bigl{(}-\lambda
X+[V,X]\bigr{)}-\frac{1}{2\lambda^{2}}\lambda[V,X]$
$\displaystyle{}=T_{X}-\dfrac{1}{2\lambda}T_{[V,X]}\in\operatorname{Ad}(g)(\mathfrak{s})\cap\mathfrak{k}.$

By Lemma 4.7, we obtain
$\operatorname{Ad}(g)Z\in\mathfrak{s}\cap\mathfrak{t}$, and thus,
$Z\in\operatorname{Ad}(g^{-1})(\mathfrak{s}\cap\mathfrak{t})=\mathfrak{s}\cap\mathfrak{t}$,
a contradiction. We conclude that $\Phi_{V}$ is the zero map for every
$V\in\mathfrak{v}_{\alpha}$.

Since $\Phi_{\mathfrak{v}_{\alpha}}\equiv 0$ we have
$[\mathfrak{s}_{\mathfrak{t}}\ominus(\mathfrak{s}\cap\mathfrak{t}),\,\mathfrak{v}_{\alpha}]=0$.
Together with $[\mathfrak{s}\cap\mathfrak{t},\mathfrak{v}_{\alpha}]=0$, this
proves $[\mathfrak{s}_{\mathfrak{t}},\mathfrak{v}_{\alpha}]=0$. Therefore, by
skew-symmetry of the elements of $\mathfrak{t}$, we get
$\langle[\mathfrak{s}_{\mathfrak{t}},\mathfrak{g}_{\alpha}\ominus\mathfrak{v}_{\alpha}],\mathfrak{v}_{\alpha}\rangle=\langle\mathfrak{g}_{\alpha}\ominus\mathfrak{v}_{\alpha},[\mathfrak{s}_{\mathfrak{t}},\mathfrak{v}_{\alpha}]\rangle=0$.
Since $\mathfrak{t}$ normalizes $\mathfrak{g}_{\alpha}$, we finally get
$[\mathfrak{s}_{\mathfrak{t}},\mathfrak{g}_{\alpha}\ominus\mathfrak{v}_{\alpha}]\subseteq\mathfrak{g}_{\alpha}\ominus\mathfrak{v}_{\alpha}$,
which in turn implies that
$\hat{\mathfrak{s}}=\mathfrak{s}_{\mathfrak{t}}\oplus\mathfrak{z}\oplus(\mathfrak{n}\ominus\mathfrak{v}_{\alpha})=\mathfrak{s}_{\mathfrak{t}}\oplus\mathfrak{s}_{\mathfrak{a}\oplus\mathfrak{n}}$
is a Lie subalgebra of $\mathfrak{t}\oplus\mathfrak{a}\oplus\mathfrak{n}$.

We can therefore consider the connected subgroup $\hat{S}$ of $G$ whose Lie
algebra is $\hat{\mathfrak{s}}$. We prove that $\hat{S}$, $S$ and $\tilde{S}$
have the same orbits. Note that a priori we do not know if $\hat{S}$ is
closed, and thus the action of $\hat{S}$ may not be proper. Since
$S\subseteq\hat{S}$ and
$\mathfrak{s}_{\mathfrak{a}\oplus\mathfrak{n}}=\hat{\mathfrak{s}}_{\mathfrak{a}\oplus\mathfrak{n}}=\mathfrak{z}\oplus(\mathfrak{n}\ominus\mathfrak{v}_{\alpha})$,
we deduce that $S\cdot o=\hat{S}\cdot o$. The same argument may be applied to
see that $\hat{S}\cdot o=\tilde{S}\cdot o$. In particular, $S\cdot
o=\hat{S}\cdot o=\tilde{S}\cdot o$ is simply connected (because $AN$ is an
exponential Lie group acting simply transitively on $M$), which means that the
isotropy subgroups $S\cap K$ and $\hat{S}\cap K$ are connected. As a
consequence, the slice representation of $\hat{S}$ at $o$ is trivial because
$[\hat{\mathfrak{s}}\cap\mathfrak{k},\mathfrak{s}_{\mathfrak{p}}^{\perp}]=[\mathfrak{s}_{\mathfrak{t}},(\mathfrak{a}\ominus\mathfrak{z})\oplus(1-\theta)\mathfrak{v}_{\alpha}]=(1-\theta)[\mathfrak{s}_{\mathfrak{t}},\mathfrak{v}_{\alpha}]=0$.
Hence, using Lemma 2.3 twice, the groups $S$, $\hat{S}$, and $\tilde{S}$ act
with the same orbits . ∎

Since we can assume that
$\mathfrak{s}_{\mathfrak{p}}^{\perp}\cap\mathfrak{p}^{1}$ has dimension one or
two, we tackle these two possibilities separately.

###  4.1. The case
$\mathfrak{s}_{\mathfrak{p}}^{\perp}\subseteq\mathfrak{p}^{1}$

Assume $\mathfrak{s}_{\mathfrak{p}}^{\perp}$ is contained in
$\mathfrak{p}^{1}$. We have that $\mathfrak{a}$ and
$\mathfrak{n}\ominus\mathfrak{n}^{1}$ are subspaces of
$\mathfrak{s}_{\mathfrak{a}\oplus\mathfrak{n}}$. A direct application of Lemma
4.3 gives $\mathfrak{n}\ominus\mathfrak{n}^{1}\subseteq\mathfrak{s}$. Let
$\xi=\sum_{\alpha\in\Lambda}(1-\theta)\xi_{\alpha}$ and
$\eta=\sum_{\alpha\in\Lambda}(1-\theta)\eta_{\alpha}$ be orthonormal vectors
in $\mathfrak{s}_{\mathfrak{p}}^{\perp}$, where $\xi_{\alpha}$,
$\eta_{\alpha}\in\mathfrak{g}_{\alpha}$. Since the action is polar
nonhyperpolar,
$[\xi,\eta]=(1+\theta)\bigl{(}\sum_{\alpha,\beta\in\Lambda}[\xi_{\alpha},\eta_{\beta}]-[\theta\xi_{\alpha},\eta_{\beta}]\bigr{)}$
is a nonzero vector orthogonal to $\mathfrak{s}$. By using the fact that
$\mathfrak{n}\ominus\mathfrak{n}^{1}\subseteq\mathfrak{s}$ and
$[\theta\xi_{\alpha},\eta_{\beta}]=0$ when $\beta\neq\alpha$, we deduce
$[\xi_{\alpha},\eta_{\alpha}]=0$, and
$[\xi_{\alpha},\eta_{\beta}]+[\xi_{\beta},\eta_{\alpha}]=0$. Thus,
$[\xi,\eta]=-(1+\theta)\sum_{\alpha\in\Lambda}[\theta\xi_{\alpha},\eta_{\alpha}]$.

Since $\mathfrak{s}_{\mathfrak{p}}^{\perp}$ is a two-dimensional Lie triple
system, it determines a totally geodesic submanifold that is isometric to a
real hyperbolic space. Hence, there exists $C>0$ such that
$\operatorname{ad}(\xi)^{2}\eta=C\eta$. Thus, we have for every
$\alpha\in\Lambda$,

$\displaystyle C\langle\xi_{\alpha},\eta_{\alpha}\rangle$
$\displaystyle{}=C\langle\xi_{\alpha},\eta\rangle=\langle\xi_{\alpha},\operatorname{ad}(\xi)^{2}\eta\rangle=\langle[\xi,\xi_{\alpha}],[\xi,\eta]\rangle$
$\displaystyle{}=-\Bigl{\langle}\sum_{\beta\in\Lambda}[\xi_{\beta},\xi_{\alpha}]-[\theta\xi_{\alpha},\xi_{\alpha}],(1+\theta)\sum_{\gamma\in\Lambda}[\theta\xi_{\gamma},\eta_{\gamma}]\Bigr{\rangle}$
$\displaystyle{}=\Bigl{\langle}\lvert\xi_{\alpha}\rvert^{2}H_{\alpha},(1+\theta)\sum_{\beta\in\Lambda}[\theta\xi_{\beta},\eta_{\beta}]\Bigr{\rangle}=0.$

######  Proposition 4.9.

$\mathfrak{s}_{\mathfrak{a}\oplus\mathfrak{n}}=(\mathfrak{a}\oplus\mathfrak{n})\ominus\mathfrak{v}_{\alpha}$,
where $\mathfrak{v}_{\alpha}$ is an abelian subspace of
$\mathfrak{g}_{\alpha}$

###### Proof.

Let $\lambda$, $\mu\in\Lambda$ with $\xi_{\lambda}$, $\xi_{\mu}\neq 0$. Since
$\langle\xi_{\alpha},\eta_{\alpha}\rangle=0$ for all simple roots
$\alpha\in\Lambda$, taking
$x=-\lvert\xi_{\lambda}\rvert^{2}/\lvert\xi_{\mu}\rvert^{2}<0$ we have
$\xi_{\lambda}+x\xi_{\mu}\in\mathfrak{s}_{\mathfrak{a}\oplus\mathfrak{n}}$.
Hence, we may find a $T\in\mathfrak{t}$ such that
$T+\xi_{\lambda}+x\xi_{\mu}\in\mathfrak{s}$. On the other hand, choose a
vector $H\in\mathfrak{a}$ with $\lambda(H)=0$ and $\mu(H)\neq 0$. Since
$\mathfrak{a}\subseteq\mathfrak{s}_{\mathfrak{a}\oplus\mathfrak{n}}$, there
exists $T^{\prime}\in\mathfrak{t}$ such that $T^{\prime}+H\in\mathfrak{s}$. As
a consequence,
$[T^{\prime},\xi_{\lambda}]+x[T^{\prime},\xi_{\mu}]+x\mu(H)\xi_{\mu}=[T^{\prime}+H,T+\xi_{\lambda}+x\xi_{\mu}]\in\mathfrak{s}$.
This means that
$0=\langle[T^{\prime},\xi_{\lambda}]+x[T^{\prime},\xi_{\mu}]+x\mu(H)\xi_{\mu},\sum_{\alpha}(1-\theta)\xi_{\alpha}\rangle=x\mu(H)\lvert\xi_{\mu}\rvert^{2}$,
which is a contradiction. Thus, $\xi=(1-\theta)\xi_{\alpha_{1}}$ for a fixed
simple root $\alpha\in\Lambda$. The same argument can be applied to conclude
that $\eta=(1-\theta)\xi_{\beta}$ for a simple root $\beta\in\Lambda$. We now
prove that $\alpha=\beta$. Indeed, if $\alpha\neq\beta$, we have
$[\xi,\eta]=(1-\theta)([\xi_{\alpha},\eta_{\beta}]-[\theta\xi_{\alpha},\eta_{\beta}])=0$,
contradicting the fact that $S$ has a non flat section.

Finally,
$\mathfrak{v}_{\alpha}=\mathbb{R}\xi_{\alpha}\oplus\mathbb{R}\eta_{\alpha}$ is
abelian by Lemma 4.5(ii). ∎

The expression obtained in Proposition 4.12 together with Proposition 4.8 with
$\mathfrak{z}=\mathfrak{a}$ imply now that the action of $S$ is orbit
equivalent to item (ii) of Theorem B.

###  4.2. The case
$\dim(\mathfrak{s}_{\mathfrak{p}}^{\perp}\cap\mathfrak{p}^{1})=1$.

In this setting, we can choose two orthonormal vectors
$\xi=\xi_{0}+\sum_{\alpha\in\Lambda}(1-\theta)\xi_{\alpha}$ and
$\eta=\sum_{\alpha\in\Lambda}(1-\theta)\eta_{\alpha}$ that span
$\mathfrak{s}_{\mathfrak{p}}^{\perp}$, where $\xi_{0}\in\mathfrak{a}$ is
nonzero and $\xi_{\alpha}$, $\eta_{\alpha}\in\mathfrak{g}_{\alpha}$ for each
$\alpha\in\Lambda$.

Since $\alpha-\beta\notin\Sigma$ for $\alpha$, $\beta\in\Lambda$,
$\alpha\neq\beta$, we have

$[\xi,\eta]=(1+\theta)\Bigl{(}\sum_{\alpha\in\Lambda}\bigl{(}\alpha(\xi_{0})\eta_{\alpha}-[\theta\xi_{\alpha},\eta_{\alpha}]\bigr{)}+\sum_{\alpha,\beta\in\Lambda}[\xi_{\alpha},\eta_{\beta}]\Bigr{)}.$

Recall that $\operatorname{ad}(\xi)^{2}\eta=C\eta$ for a positive constant
$C\in\mathbb{R}$. In particular, for any $H\in\mathfrak{a}$,

$\displaystyle 0$ $\displaystyle{}=\langle
C\eta,\,H\rangle=\langle\operatorname{ad}(\xi)^{2}\eta,\,H\rangle=\langle[\xi,\eta],\,[\xi,H]\rangle=-\Bigl{\langle}[\xi,\eta],\,(1+\theta)\sum_{\gamma\in\Lambda}\gamma(H)\xi_{\gamma}\Bigr{\rangle}$
$\displaystyle{}=-2\sum_{\alpha\in\Lambda}\alpha(\xi_{0})\alpha(H)\langle\xi_{\alpha},\eta_{\alpha}\rangle=\Bigl{\langle}-2\sum_{\alpha\in\Lambda}\alpha(\xi_{0})\langle\xi_{\alpha},\eta_{\alpha}\rangle
H_{\alpha},\,H\Bigr{\rangle},$

which implies
$-2\sum_{\alpha\in\Lambda}\alpha(\xi_{0})\langle\xi_{\alpha},\eta_{\alpha}\rangle
H_{\alpha}=0$. Therefore,

(3) $\alpha(\xi_{0})\langle\xi_{\alpha},\eta_{\alpha}\rangle=0,\text{ for
every $\alpha\in\Lambda$.}$

On the other hand, since $\operatorname{ad}(\eta)^{2}\xi=C\xi$, for any
$H\in\mathfrak{a}$,

$\displaystyle C\langle\xi_{0},H\rangle$ $\displaystyle{}=\langle
C\xi,H\rangle=\langle\operatorname{ad}(\eta)^{2}\xi,\,H\rangle=\langle[\eta,\xi],\,[\eta,H]]\rangle=\Bigl{\langle}[\xi,\eta],\,(1+\theta)\sum_{\gamma\in\Lambda}\gamma(H)\eta_{\gamma}\Bigr{\rangle}$
$\displaystyle{}=2\sum_{\alpha\in\Lambda}\alpha(\xi_{0})\alpha(H)\lvert\eta_{\alpha}\rvert^{2}=\Bigl{\langle}2\sum_{\alpha\in\Lambda}\alpha(\xi_{0})\lvert\eta_{\alpha}\rvert^{2}H_{\alpha},\,H\Bigr{\rangle},$

we obtain

(4)
$C\xi_{0}=2\sum_{\alpha\in\Lambda}\alpha(\xi_{0})\lvert\eta_{\alpha}\rvert^{2}H_{\alpha}.$

Similarly, let $\alpha,\beta\in\Lambda$ be arbitrary, and
$X\in\mathfrak{g}_{\alpha+\beta}$. Then,

$\displaystyle 0$ $\displaystyle{}=\langle
C\xi,X\rangle=\langle\operatorname{ad}(\eta)^{2}\xi,\,X\rangle=\langle[\eta,\xi],\,[\eta,X]\rangle=-\Bigl{\langle}[\xi,\eta],\,\sum_{\mu\in\Lambda}\bigl{(}[X,\eta_{\mu}]-[X,\theta\eta_{\mu}]\bigr{)}\Bigr{\rangle}$
$\displaystyle{}=-\Bigl{\langle}\sum_{\gamma\in\Lambda}\gamma(\xi_{0})\eta_{\gamma},\,\sum_{\delta\in\Lambda}[X,\theta\eta_{\delta}]\Bigr{\rangle}=\sum_{\gamma,\delta\in\Lambda}\gamma(\xi_{0})\langle[\eta_{\gamma},\eta_{\delta}],\,X\rangle$
$\displaystyle{}=\langle\alpha(\xi_{0})[\eta_{\alpha},\eta_{\beta}]+\beta(\xi_{0})[\eta_{\beta},\eta_{\alpha}],X\rangle.$

Consequently, for any two simple roots $\alpha,\beta\in\Lambda$,
$[(\alpha-\beta)(\xi_{0})\,\eta_{\alpha},\eta_{\beta}]=0$.

######  Lemma 4.10.

We have $\langle\xi_{\alpha},\eta_{\alpha}\rangle=0$ for all
$\alpha\in\Lambda$.

###### Proof.

We define $\Psi=\\{\alpha\in\Lambda\colon\eta_{\alpha}\neq 0\\}$. We show that
$\langle\xi_{\alpha},\eta_{\alpha}\rangle=0$ for each $\alpha\in\Psi$.

From (4) the map $\alpha\in\Psi\mapsto\alpha(\xi_{0})$ cannot be identically
zero. Thus, fix $\alpha\in\Psi$ such that $\alpha(\xi_{0})\neq 0$. Hence (3)
already implies $\langle\xi_{\alpha},\eta_{\alpha}\rangle=0$.

Assume $\beta\in\Psi$ satisfies $\langle\xi_{\beta},\eta_{\beta}\rangle\neq
0$. In particular, from (3) we have $\beta(\xi_{0})=0$. If
$\langle\alpha,\beta\rangle\neq 0$, the linear map
$\operatorname{ad}(\eta_{\beta})\colon\mathfrak{g}_{\alpha}\to\mathfrak{g}_{\alpha+\beta}$
is injective. From $[(\alpha-\beta)(\xi_{0})\,\eta_{\alpha},\eta_{\beta}]=0$
we deduce $(\alpha-\beta)(\xi_{0})\eta_{\alpha}=0$, so
$\alpha(\xi_{0})=\beta(\xi_{0})=0$, contradiction. Thus,
$\langle\alpha,\beta\rangle=0$. In particular,
$H_{\beta}\in\mathfrak{s}_{\mathfrak{a}\oplus\mathfrak{n}}$, so there exists
$T\in\mathfrak{t}$ such that $T+H_{\beta}\in\mathfrak{s}$. On the other hand,
$\xi_{0}+x\eta_{\alpha}+y\eta_{\beta}\in\mathfrak{s}_{\mathfrak{a}\oplus\mathfrak{n}}$
for $y=-\lvert\xi_{0}\rvert^{2}/\langle\xi_{\beta},\eta_{\beta}\rangle\neq 0$
and $x=-y\lvert\eta_{\beta}\rvert^{2}/\lvert\eta_{\alpha}\rvert^{2}\neq 0$, so
$T^{\prime}+\xi_{0}+x\eta_{\alpha}+y\eta_{\beta}\in\mathfrak{s}$ for an
adequate $T^{\prime}\in\mathfrak{t}$. Thus,

$\displaystyle 0$
$\displaystyle{}=\langle[T+H_{\beta},T^{\prime}+\xi_{0}+x\eta_{\alpha}+y\eta_{\beta}],\,\eta\rangle$
$\displaystyle{}=\langle
x[T,\eta_{\alpha}]+y[T,\eta_{\beta}]+y\lvert\beta\rvert^{2}\eta_{\beta},\,\eta\rangle=y\lvert\beta\rvert^{2}\lvert\eta_{\beta}\rvert^{2},$

which gives us a contradiction. Therefore,
$\langle\xi_{\beta},\eta_{\beta}\rangle=0$ for all $\beta\in\Lambda$. ∎

######  Proposition 4.11.

There exists a simple root $\alpha\in\Lambda$ and a constant $a\in\mathbb{R}$
such that $\xi=aH_{\alpha}+(1-\theta)\xi_{\alpha}$. If $\xi_{\alpha}=0$ (that
is, if $\xi\in\mathfrak{a}$), then $\eta=(1-\theta)\eta_{\alpha}$.

###### Proof.

Firstly, suppose $\xi\in\mathfrak{a}$. Then, a direct application of Lemma 4.5
implies that $\eta=(1-\theta)\eta_{\alpha}$ for a simple root
$\alpha\in\Lambda$. In particular, (4) is reduced to
$C\xi_{0}=2\alpha(\xi_{0})\lvert\eta_{\alpha}\rvert^{2}H_{\alpha}$, so
$\xi_{0}\in\mathbb{R}H_{\alpha}$, and the proposition follows.

Now, assume $\xi_{0}\notin\mathfrak{a}$, and let $\alpha\in\Lambda$ be a
simple root such that $\xi_{\alpha}\neq 0$. We prove that
$\xi=aH_{\alpha}+(1-\theta)\xi_{\alpha}$.

Suppose that $\xi_{0}$ is not proportional to $H_{\alpha}$. Then there exists
$H\in\mathfrak{a}$ such that $\langle H,\xi_{0}\rangle=0$ and $\alpha(H)\neq
0$. As a consequence, $H\in\mathfrak{s}_{\mathfrak{a}\oplus\mathfrak{n}}$, and
there exists $T\in\mathfrak{t}$ for which $T+H\in\mathfrak{s}$. On the other
hand, $\xi_{0}+x\xi_{\alpha}\in\mathfrak{s}_{\mathfrak{a}\oplus\mathfrak{n}}$
for $x=-\lvert\xi_{0}\rvert^{2}/\lvert\xi_{\alpha}\rvert^{2}<0$ (because
$\langle\xi_{\alpha},\eta_{\alpha}\rangle=0$ by Lemma 4.10), and we may choose
$T^{\prime}\in\mathfrak{t}$ such that
$T^{\prime}+\xi_{0}+x\xi_{\alpha}\in\mathfrak{s}$. Thus,
$[T+H,T^{\prime}+\xi_{0}+x\xi_{\alpha}]=x[T,\xi_{\alpha}]+x\alpha(H)\xi_{\alpha}\in\mathfrak{s}$.
Taking the inner product with $\xi$ yields
$x\alpha(H)\lvert\xi_{\alpha}\rvert^{2}=0$, a contradiction.

Hence $\xi_{0}\in\mathbb{R}H_{\alpha}$ for any $\alpha\in\Lambda$. The fact
that simple roots are linearly independent together with (4) implies
$\xi_{\beta}=0$ for every $\beta\in\Lambda\setminus\\{\alpha\\}$. ∎

So far we have proved that $\xi$ must take the form
$aH_{\alpha}+(1-\theta)\xi_{\alpha}$ for a nonzero $a\in\mathbb{R}$ and
$\xi_{\alpha}\in\mathfrak{g}_{\alpha}$ (which may be zero). If
$\xi_{\alpha}=0$, then we also know that $\eta=(1-\theta)\eta_{\alpha}$. If
$\xi_{\alpha}\neq 0$, then the third statement of Lemma 4.5 implies that the
action of $S$ is orbit equivalent to an action of another closed connected
subgroup $\tilde{S}$ for which the normal space of $\tilde{S}\cdot o$ at $o$
takes the form
$\tilde{\mathfrak{s}}_{\mathfrak{p}}^{\perp}=\\{\xi_{\alpha},bH_{\alpha}+(1-\theta)\nu_{\alpha}\\}$
for a constant $b\in\mathbb{R}$ and a $\nu_{\alpha}\in\mathfrak{g}_{\alpha}$.
Because of this, we may assume without loss of generality that
$\mathfrak{s}_{\mathfrak{p}}^{\perp}$ is spanned by two orthogonal vectors
$\xi=aH_{\alpha}+(1-\theta)\xi_{\alpha}$, $\eta=(1-\theta)\eta_{\alpha}$, with
$a\neq 0$, $\xi_{\alpha}$, $\eta_{\alpha}\in\mathfrak{g}_{\alpha}$. Recall
from Lemma 4.5 that $[\xi_{\alpha},\eta_{\alpha}]=0$.

The key to finishing the proof lies in the following result:

######  Proposition 4.12.

Assume
$\mathfrak{s}_{\mathfrak{p}}^{\perp}=\operatorname{span}\\{\xi,\eta\\}$, where
$\xi=aH_{\alpha}+(1-\theta)\xi_{\alpha}$ and $\eta=(1-\theta)\eta_{\alpha}$,
where $a\neq 0$, $\xi_{\alpha}$, $\eta_{\alpha}\in\mathfrak{g}_{\alpha}$ are
orthogonal commuting vectors, and $\alpha\in\Lambda$. Then:

  1. (i)

If $\xi_{\alpha}=0$, then the action of $S$ has the same orbits as the action
of the connected subgroup of $G$ whose Lie algebra is
$(\mathfrak{a}\ominus\mathbb{R}H_{\alpha})\oplus(\mathfrak{n}\ominus\mathbb{R}\eta_{\alpha})$.

  2. (ii)

If $\xi_{\alpha}\neq 0$, then there exists an abelian subspace
$\mathfrak{v}_{\alpha}\subseteq\mathfrak{g}_{\alpha}$ such that the action of
$S$ is orbit equivalent to the action of the connected subgroup of $G$ whose
Lie algebra is $\mathfrak{a}\oplus(\mathfrak{n}\ominus\mathfrak{v}_{\alpha})$.

###### Proof.

If $\xi_{\alpha}=0$, then
$\mathfrak{s}_{\mathfrak{p}}^{\perp}=\mathbb{R}H_{\alpha}\oplus\mathbb{R}(1-\theta)\eta_{\alpha}$
and
$\mathfrak{s}_{\mathfrak{a}\oplus\mathfrak{n}}=(\mathfrak{a}\ominus\mathbb{R}H_{\alpha})\oplus(\mathfrak{n}\ominus\mathbb{R}\eta_{\alpha})$.
Then, statement (i) follows directly from Proposition 4.8.

We prove (ii). We consider the element
$g=\operatorname{Exp}(-\frac{a}{\lvert\xi_{\alpha}\rvert^{2}}\xi_{\alpha})\in
N$. Since $\mathfrak{s}$ is orthogonal to $aH_{\alpha}+\xi_{\alpha}$ and
$\eta_{\alpha}$, it follows that $\operatorname{Ad}(g)\mathfrak{s}$ is
orthogonal to $\operatorname{Ad}(g^{-1})^{*}(aH_{\alpha}+\xi_{\alpha})$ and
$\operatorname{Ad}(g^{-1})^{*}\eta_{\alpha}$. By direct computation,

$\displaystyle\operatorname{Ad}(g^{-1})^{*}(aH_{\alpha}+\xi_{\alpha})$
$\displaystyle{}=e^{-\frac{a}{\lvert\xi_{\alpha}\rvert^{2}}\operatorname{ad}(\theta\xi_{\alpha})}(aH_{\alpha}+\xi_{\alpha})\equiv\xi_{\alpha}\pmod{\theta\mathfrak{n}},$
$\displaystyle\operatorname{Ad}(g^{-1})^{*}\eta_{\alpha}$
$\displaystyle{}=e^{-\frac{a}{\lvert\xi_{\alpha}\rvert^{2}}\operatorname{ad}(\theta\xi_{\alpha})}\eta_{\alpha}\equiv\eta_{\alpha}-\frac{a}{\lvert\xi_{\alpha}\rvert^{2}}[\theta\xi_{\alpha},\eta_{\alpha}]\pmod{\theta\mathfrak{n}},$

and since
$\operatorname{Ad}(g)\mathfrak{s}\subseteq\mathfrak{t}\oplus\mathfrak{a}\oplus\mathfrak{n}$,
it follows that the vectors $\xi_{\alpha}$ and
$\eta_{\alpha}-\frac{a}{\lvert\xi_{\alpha}\rvert^{2}}[\theta\xi_{\alpha},\eta_{\alpha}]$
are orthogonal to $\operatorname{Ad}(g)\mathfrak{s}$. On the other hand, the
action of $S$ is polar, so
$[\xi,\eta]=a\lvert\alpha\rvert^{2}(1+\theta)\eta_{\alpha}-2[\theta\xi_{\alpha},\eta_{\alpha}]$
is also orthogonal to $\mathfrak{s}$. As a consequence, $\langle
a\lvert\alpha\rvert^{2}\eta_{\alpha}-2[\theta\xi_{\alpha},\eta_{\alpha}],\mathfrak{s}\rangle=0$.
We deduce that

$\operatorname{Ad}(g^{-1})^{*}(a\lvert\alpha\rvert^{2}\eta_{\alpha}-2[\theta\xi_{\alpha},\eta_{\alpha}])\equiv
a\lvert\alpha\rvert^{2}\eta_{\alpha}-\Bigl{(}2+\frac{a^{2}\lvert\alpha\rvert^{2}}{\lvert\xi_{\alpha}\rvert^{2}}\Bigr{)}[\theta\xi_{\alpha},\eta_{\alpha}]\pmod{\theta\mathfrak{n}}$

is also orthogonal to $\operatorname{Ad}(g)\mathfrak{s}$. Because
$\theta\mathfrak{n}$ is already orthogonal to $\mathfrak{s}$, it follows that
$a\lvert\alpha\rvert^{2}\eta_{\alpha}-\left(2+\frac{a^{2}\lvert\alpha\rvert^{2}}{\lvert\xi_{\alpha}\rvert^{2}}\right)[\theta\xi_{\alpha},\eta_{\alpha}]$
is perpendicular to $\mathfrak{s}$. Since

$\begin{vmatrix}1&-\frac{a}{\lvert\xi_{\alpha}\rvert^{2}}\\\
a\lvert\alpha\rvert^{2}&-2-\frac{a^{2}\lvert\alpha\rvert^{2}}{\lvert\xi_{\alpha}\rvert^{2}}\end{vmatrix}=-2<0,$

we deduce that $\eta_{\alpha}$ and $[\theta\xi_{\alpha},\eta_{\alpha}]$ are
both orthogonal to $\mathfrak{s}$. We conclude that
$(\operatorname{Ad}(g)\mathfrak{s})_{\mathfrak{p}}^{\perp}=(1-\theta)\mathfrak{v}_{\alpha}$,
where
$\mathfrak{v}_{\alpha}=\operatorname{span}\\{\xi_{\alpha},\eta_{\alpha}\\}$,
and thus, $gSg^{-1}$ and the connected subgroup of $G$ whose Lie algebra is
$\mathfrak{a}\oplus(\mathfrak{n}\ominus\mathfrak{v}_{\alpha})$ act with the
same orbits due to Proposition 4.8. ∎

The proof of Theorem B follows now from the observation that Proposition
4.12(i) corresponds to case (i), and Proposition 4.12(ii) corresponds to case
(ii).

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