Symmetric spaces with dissecting involutions

TL;DR

This paper classifies all irreducible symmetric spaces with dissecting involutions, revealing that only spheres and hyperbolic spaces admit such involutions with fixed point spaces of one dimension lower.

Contribution

It provides a complete classification of dissecting involutions on irreducible symmetric spaces, extending understanding beyond Riemannian cases.

Findings

Only spheres and hyperbolic spaces have dissecting involutions among irreducible 1-connected Riemannian symmetric spaces.

Dissecting involutions have fixed point spaces of codimension one, namely spheres and hyperbolic spaces.

The classification links dissecting involutions to geometric structures relevant in quantum field theory.

Abstract

An involutive diffeomorphism $\sigma$ of a connected smooth manifold $M$ is called dissecting if the complement of its fixed point set is not connected. Dissecting involutions on a complete Riemannian manifold are closely related to constructive quantum field theory through the work of Dimock and Jaffe/Ritter on the construction of reflection positive Hilbert spaces. In this article we classify all pairs $(M,\sigma)$, where $M$ is an irreducible symmetric space, not necessarily Riemannian, and $\sigma$ is a dissecting involutive automorphism. In particular, we show that the only irreducible $1$-connected Riemannian symmetric spaces are $S^n$ and $H^n$ with dissecting isometric involutions whose fixed point spaces are $S^{n-1}$ and $H^{n-1}$, respectively.

Equations

$$\dim ({\mathfrak{g}}^{-\tau }\cap {\mathfrak{g}}^{-\sigma })=1.$$

$$Q:=\{x\in {\mathbb{R}}^{p+q}:{\beta }_{p,q}(x,x)=1\},\text{mbox}where{\beta }_{p,q}(x,y)=\sum _{j=1}^p{x}_j{y}_j-\sum _{j=p+1}^{p+q}{x}_j{y}_j.$$

$$s_x(y\cdot z)=s_x(y)\cdot s_x(z)\text{mbox}forallx,y\in M.$$

$$\dim ({\mathfrak{g}}^{-\tau }\cap {\mathfrak{g}}^{-\sigma })=1.$$

$${\mathbb{P}}^n={\mathbb{S}}^n/\{\pm 1\}\cong {\mathrm{SO}}_{n+1}(\mathbb{R})/\mathrm{S}({\mathrm{O}}_1(\mathbb{R})\times {\mathrm{O}}_n(\mathbb{R})).$$

$$\sigma ([{\xi }_1,\dots ,{\xi }_{n+1}])=[-{\xi }_1,{\xi }_2,\dots ,{\xi }_{n+1}]=[{\xi }_1,-{\xi }_2,\dots ,-{\xi }_{n+1}]$$

$$M_{+}/\{\pm 1\}=M_{-}/\{\pm 1\}=\{[{\xi }_1,\dots ,{\xi }_{n+1}]:{\xi }_1\ne 0\},$$

$$[\mathfrak{h},{\mathfrak{j}}_{\mathfrak{h}}]\subseteq {\mathfrak{j}}_{\mathfrak{h}},[\mathfrak{h},{\mathfrak{j}}_{\mathfrak{q}}]\subseteq {\mathfrak{j}}_{\mathfrak{q}},[\mathfrak{q},{\mathfrak{j}}_{\mathfrak{h}}]\subseteq {\mathfrak{j}}_{\mathfrak{q}},[\mathfrak{q},{\mathfrak{j}}_{\mathfrak{q}}]\subseteq {\mathfrak{j}}_{\mathfrak{h}},$$

$${\sigma }_x(y):=y-2\frac{\beta (x,y)}{\beta (x,x)}x$$

$$Q_c:=Q_c(V,\beta ):=\{v\in V:\beta (v,v)=c\}$$

$$M_{\pm }:=\{x\in M:\pm \beta (x,y)>0\}$$

$${\beta }_{p,q}(x,y)=\sum _{j=1}^p{x}_j{y}_j-\sum _{j=p+1}^{p+q}{x}_j{y}_j.$$

$${\mathfrak{g}}^c\cong \{\begin{array}{ll}{\mathrm{so}}_{p-1,q+1}(\mathbb{R})& \text{ for }j\le p\\ {\mathrm{so}}_{p+1,q-1}(\mathbb{R})& \text{ for }j>p.\end{array}$$

$$\beta (X,Y):=\frac{1}{2}(\mathrm{tr}(X)\mathrm{tr}(Y)-\mathrm{tr}(XY)).$$

$$M:={\mathrm{SL}}_2(\mathbb{R})=\{X\in M_2(\mathbb{R}):\det (X)=1\}$$

$${\sigma }_1^{\mathfrak{g}}(x)=x^{\top }\text{mbox}and{\sigma }_2^{\mathfrak{g}}\left(\begin{array}{cc}a& b\\ c& -a\end{array}\right)=\left(\begin{array}{cc}-a& b\\ c& a\end{array}\right)=-I_{1,1}x{I}_{1,1}\text{mbox}for{I}_{1,1}=\left(\begin{array}{cc}1& 0\\ 0& -1\end{array}\right).$$

$${\sigma }_1^G(g)=g^{\top }\text{mbox}and{\sigma }_2^G(g)=I_{1,1}{g}^{-1}{I}_{1,1}.$$

$${\sigma }_1(x,y)=(-y^{\top },-x^{\top })\text{mbox}and{\sigma }_2(x,y)=(I_{1,1}y{I}_{1,1},I_{1,1}x{I}_{1,1}).$$

$$\beta (x,y):=\frac{1}{2}\mathrm{tr}(x{y}^{\ast }).$$

$$M:={\mathrm{SU}}_2(\mathbb{C})=\{x\in \mathbb{H}:\beta (x,x)=\det (x)=1\}$$

$${\sigma }^{\mathfrak{g}}\left(\begin{array}{cc}ai& b\\ -b^{\ast }& -ai\end{array}\right)=\left(\begin{array}{cc}-ai& b\\ -b^{\ast }& ai\end{array}\right)=-I_{1,1}x{I}_{1,1}.$$

$$\sigma (x,y)=(I_{1,1}y{I}_{1,1},I_{1,1}x{I}_{1,1}).$$

$$\mathfrak{g}=\mathfrak{h}\oplus \mathfrak{q}=\mathfrak{l}\oplus \mathfrak{m}={\mathfrak{h}}_{\mathfrak{l}}\oplus {\mathfrak{h}}_{\mathfrak{m}}\oplus {\mathfrak{q}}_{\mathfrak{l}}\oplus {\mathfrak{q}}_{\mathfrak{m}}$$

$$({\mathfrak{g}}^r{)}^{-{\tau }^r}\cap ({\mathfrak{g}}^r{)}^{-{\sigma }^r}=(\mathfrak{q}\cap \mathfrak{k}\oplus i(\mathfrak{q}\cap \mathfrak{p}))\cap (\mathfrak{m}\cap \mathfrak{k}\oplus i(\mathfrak{m}\cap \mathfrak{p}))=({\mathfrak{q}}_{\mathfrak{m}}\cap \mathfrak{k})\oplus i({\mathfrak{q}}_{\mathfrak{m}}\cap \mathfrak{p}).$$

$${\mathfrak{g}}^c=\{(x,x)+i(iy,-iy)=(x-y,x+y):x,y\in {\mathfrak{g}}_1\}\cong {\mathfrak{g}}_1\oplus {\mathfrak{g}}_1\text{mbox}with{\tau }^c(z,w)=(w,z).$$

$$\sigma (x,y)=({\sigma }_1(x),{\sigma }_1(y))\text{mbox}or\sigma (x,y)=({\sigma }_1(y),{\sigma }_1(x))\text{mbox}forx,y\in \mathfrak{g}.$$

$$\sigma (x,x)=({\sigma }_1(x),{\sigma }_1(x))\text{mbox}forx\in {\mathfrak{g}}_1.$$

$$\gamma (x,-x)=(c(x),-c(x))\text{mbox}forx\in {\mathfrak{g}}_1.$$

$$c([x,y])=[x,c(y)]=[c(x),y]\text{mbox}forx,y\in {\mathfrak{g}}_1.$$

$$([c(x),c(y)],[c(x),c(y)])$$

Full text

The research of K.-H. Neeb was partially supported by DFG-grant NE 413/9-1. The research of G. Ólafsson was partially supported by Simons grant 586106.

1. Introduction

In this article we classify dissecting involutions on irreducible symmetric spaces. An involutive diffeomorphism $\sigma$ of a connected smooth manifold $M$ is called dissecting if the complement of the fixed point manifold $M^{\sigma}$ is not connected. Then $M\setminus M^{\sigma}$ has two connected components $M_{\pm}$ exchanged by $\sigma$ and $\mathop{{\rm codim}}\nolimits M^{\sigma}=1$.

We are interested in the situation where $M$ is a symmetric space and $\sigma$ is an automorphism of $M$. Any connected symmetric space is of the form $M\cong G/H$, where $G$ is a connected Lie group acting faithfully by automorphisms on $M$ and there exists an involutive automorphism $\tau$ of $G$ for which $H$ is an open subgroup of $G^{\tau}=\{g\in G\colon\tau(g)=g\}$, the subgroup of $\tau$-fixed points. Then $\tau$ induces an involutive automorphism, also denoted $\tau$, on the Lie algebra ${\mathfrak{g}}$ of $G$, so that we obtain the symmetric Lie algebra $({\mathfrak{g}},\tau)$. Any dissecting involutive automorphism $\sigma_{M}$ of $M$, fixing the base point $eH\in M$, leads to another involutive automorphism $\sigma$ of ${\mathfrak{g}}$, commuting with $\tau$, such that

[TABLE]

For a simply connected symmetric space $M=G/H$, this condition is equivalent to $\sigma_{M}$ being dissecting. Accordingly, we call a triple $({\mathfrak{g}},\tau,\sigma)$, consisting of a Lie algebra and two commuting involutive automorphisms $\tau$ and $\sigma$, dissecting if (1.1) is satisfied.

In this note we classify all dissecting triples $({\mathfrak{g}},\tau,\sigma)$ which are irreducible in the sense that ${\mathfrak{g}}$ contains no non-trivial $\tau$-invariant ideals. On the global level, this classifies connected, simply connected symmetric spaces with dissecting involutions (Theorem 2.4). For any connected symmetric space $M$ with a dissecting involution $\sigma_{M}$, the canonical lift $\sigma_{\widetilde{M}}$ of $\sigma_{M}$ to the simply connected covering $\widetilde{M}$ is also dissecting, but the converse is in general not true (cf. Example 2.6). The classification is easy to state:

Theorem 1.1**.**

Up to coverings connected irreducible symmetric space with a dissecting involutive automorphism is a connected component of a quadric

[TABLE]

Here $G=\mathop{{\rm SO}}\nolimits_{p,q}({\mathbb{R}})_{0}$ and $H_{0}\cong\mathop{{\rm SO}}\nolimits_{p-1,q}({\mathbb{R}})_{0}$ or $\mathop{{\rm SO}}\nolimits_{p,q-1}({\mathbb{R}})_{0}$. Up to conjugation, the dissecting involution is given in the first case by $\sigma(x_{1},\ldots,x_{p+q})=(-x_{1},x_{2},\ldots,x_{p+q})$ and in the second case by $\sigma(x_{1},\ldots,x_{p+q})=(x_{1},\ldots,x_{n-1},-x_{p+q})$.

Our motivation to classify symmetric spaces with dissecting involutions comes from constructions in Quantum Field Theory and our work on reflection positivity, see [NÓ18] and the references therein. Starting from a pair $(M,\sigma)$ of a complete Riemannian manifold $M$ and a dissecting involutive isometry one produces a reflection positive Hilbert space [Di04, JR08, AFG86, NÓ18] which provides the basis for constructing relativistic field theories on a Lorentzian manifold. Typical examples arise in physics from isometric dissecting involutions on flat euclidean space ${\mathbb{R}}^{n}$, the (positively curved) sphere ${\mathbb{S}}^{n}$ and the (negatively curved) hyperbolic space ${\mathbb{H}}^{n}$. Our classification implies in particular that these three types cover all (irreducible) Riemannian symmetric spaces with dissecting isometries. The special case of $M={\mathbb{S}}^{n}$ and related questions of harmonic analysis and representation theory is explored in [NÓ19].

The article is organized as follows. In Section 2 we introduce the basic concepts. In Section 3 we discuss several examples of homogeneous spaces with dissecting involutions, and Section 4 is devoted to the proof of the classification theorem.

2. Dissecting involutions on manifolds and symmetric spaces

In this section we introduce the two basic concepts discussed in this note, dissections and symmetric spaces.

2.1. Dissections on manifolds

Definition 2.1**.**

(a) A diffeomorphism $\sigma$ of a connected smooth manifold $M$ is called dissecting if the complement of the fixed point set $M^{\sigma}$ is not connected.

(b) A smooth involution $\sigma\colon M\to M$ is called a reflection if there exists a fixed point $p\in M^{\sigma}$ such that the fixed point space of the linear involution $T_{p}(\sigma)$ on $T_{p}(M)$ is a hyperplane.

Theorem 2.2**.**

Let $M$ be a connected smooth manifold.

• (i)

If $g$ is a complete Riemannian metric on $M$ and $\sigma$ is a dissecting isometry, then $\sigma$ is an involution.

• (ii)

If $\sigma\in\mathop{{\rm Diff}}\nolimits(M)$ is an involution, then there exists a $\sigma$-invariant complete Riemannian metric on $M$.

• (iii)

If $\sigma\in\mathop{{\rm Diff}}\nolimits(M)$ is a dissecting involution, then $M\setminus M^{\sigma}$ has two connected components $M_{\pm}$ with $\sigma(M_{\pm})=M_{\mp}$ and each component of $M^{\sigma}$ is of codimension one.

• (iv)

If $\sigma\in\mathop{{\rm Diff}}\nolimits(M)$ is a reflection and $M$ is simply connected, then $\sigma$ is dissecting and its fixed point set $M^{\sigma}$ is a connected orientable hypersurface.

Proof.

(i) follows from [AKLM06, Lemma 2.7].

(ii) From [NO61] we obtain a complete Riemannian metric $g$ on $M$. Then $\widetilde{g}\mathrel{\mathop{\ordinarycolon}}=g+\sigma^{*}g$ is also complete by the Hopf–Rinow Theorem because its closed balls are contained in closed $g$-balls, hence compact.

(iii), (iv): By (ii), we can assume that $\sigma$ is isometric with respect to a complete Riemannian metric. Hence (iii) follows from [AKLM06, Lemma 2.7] and (iv) from [AKLM06, Thm. 2.8]. ∎

2.2. Symmetric spaces

Definition 2.3**.**

(a) Let $M$ be a smooth manifold and $\mu\colon M\times M\to M,(x,y)\mapsto x\cdot y=\mathrel{\mathop{\ordinarycolon}}s_{x}(y)$ be a smooth map with the following properties: each $s_{x}$ is an involution for which $x$ is an isolated fixed point and

[TABLE]

Then we call $(M,\mu)$ a symmetric space.

(b) A morphism of symmetric spaces $M$ and $N$ is a smooth map $\varphi\mathrel{\mathop{\ordinarycolon}}M\to N$ such that $\varphi(x\cdot y)=\varphi(x)\cdot\varphi(y)$ for $x,y\in M$.

As shown in [Lo69], connected symmetric spaces are homogeneous spaces of Lie groups and they arise from the following construction, which goes back to É. Cartan (see [Hel78] for a detailed discussion). Let $G$ be a Lie group and $\tau\mathrel{\mathop{\ordinarycolon}}G\to G$ be an involutive automorphism. If $H\subset G^{\tau}$ is an open subgroup, then the homogeneous space $M\mathrel{\mathop{\ordinarycolon}}=G/H$ is a symmetric space with respect to the involutions $s_{gH}(xH)\mathrel{\mathop{\ordinarycolon}}=g\tau(g^{-1}x)H$. Then $\tau$ defines an involution of ${\mathfrak{g}}$, also denoted by $\tau$. We have a decomposition of ${\mathfrak{g}}$ into eigenspaces ${\mathfrak{g}}={\mathfrak{h}}\oplus{\mathfrak{q}}$ where ${\mathfrak{h}}={\mathfrak{g}}^{\tau}$ and ${\mathfrak{q}}={\mathfrak{g}}^{-\tau}$. Here ${\mathfrak{h}}$ is the Lie algebra of $H$, and ${\mathfrak{q}}$ is identified with the tangent space $T_{eH}(G/H)$.

Let $(M,(s_{x})_{x\in M})$ be a connected symmetric space and $\sigma_{M}$ a dissecting involutive automorphism of $M$. Then the subset $M^{\sigma_{M}}$ of fixed points is non-empty and we may pick $x\in M^{\sigma_{M}}$. Writing $M\cong G/H$ for a connected transitive Lie group $G$ of automorphisms and the stabilizer group $H\mathrel{\mathop{\ordinarycolon}}=G_{x}$, the reflection $s_{x}$ induces an involutive automorphism $\tau$ on the Lie algebra ${\mathfrak{g}}$ of $G$, and $\sigma_{M}(x)=x$ implies that $\sigma_{M}$ commutes with $s_{x}$ by (2.1), so that it induces an involution $\sigma$ on ${\mathfrak{g}}$ commuting with $\tau$. That $\sigma_{M}$ is dissecting implies that $({\mathfrak{g}}^{-\tau})^{\sigma}$ is a hyperplane in ${\mathfrak{g}}^{-\tau}\cong T_{x}(M)$. This proves the “only if” direction in the following theorem.

Theorem 2.4**.**

Let $M=G/H$ be a simply connected symmetric space and $\sigma_{M}$ an involutive automorphism of $M$ leaving the base point $x=eH$ invariant. Denote the corresponding involution on ${\mathfrak{g}}$ by $\sigma$. Then $\sigma_{M}$ is dissecting if and only if $\mathop{{\rm dim}}\nolimits({\mathfrak{g}}^{-\sigma}\cap{\mathfrak{g}}^{-\tau})=1$.

Proof.

If $\sigma_{M}$ is dissecting, then we have just seen that $\mathop{{\rm dim}}\nolimits({\mathfrak{g}}^{-\sigma}\cap{\mathfrak{g}}^{-\tau})=1$. The other direction follows from Theorem 2.2(iv) because the involution $T_{eH}(\sigma_{M})$ on $T_{eH}(M)$ corresponds to the involution $\sigma|_{{\mathfrak{q}}}$ whose fixed point space ${\mathfrak{q}}^{\sigma}={\mathfrak{g}}^{-\tau}\cap{\mathfrak{g}}^{\sigma}$ is a hyperplane. ∎

Definition 2.5**.**

We call a triple $({\mathfrak{g}},\tau,\sigma)$, consisting of a real Lie algebra ${\mathfrak{g}}$ and two commuting involutions $\tau,\sigma\in\mathop{{\rm Aut}}\nolimits({\mathfrak{g}})$ dissecting if

[TABLE]

A triple $({\mathfrak{g}},\tau,\sigma)$ is called semisimple if ${\mathfrak{g}}$ is semisimple. We say that it is irreducible if $({\mathfrak{g}},\tau)$ is irreducible, i.e., ${\mathfrak{g}}^{-\tau}\not=\{0\}$, and $\{0\}$ and ${\mathfrak{g}}$ are the only $\tau$-invariant ideals of ${\mathfrak{g}}$.

Example 2.6**.**

We now show that a dissecting triple $({\mathfrak{g}},\tau,\sigma)$ may also correspond to symmetric spaces $G/H$ on which the involution induced by $\sigma$ is not dissecting. By Theorem 2.4 this can only happen if $G/H$ is not simply connected.

For $n\geq 1$ and $e_{1},\ldots,e_{n+1}$ the standard base for ${\mathbb{R}}^{n+1}$ we denote by $r_{j}$ the reflection $r_{j}(x)=x-2x_{j}e_{j}$. The $n$-dimensional sphere $M={\mathbb{S}}^{n}\cong\mathop{{\rm SO}}\nolimits_{n+1}({\mathbb{R}})e_{1}\cong\mathop{{\rm SO}}\nolimits_{n+1}({\mathbb{R}})/\mathop{{\rm SO}}\nolimits_{n}({\mathbb{R}})$ is a connected symmetric space in which the base point $e_{1}$ corresponds to the involution $\tau(x)=r_{1}xr_{1}$. At the same time, the involution $\sigma\mathrel{\mathop{\ordinarycolon}}=r_{1}$ is dissecting with fixed point space $M^{\sigma}=\{x\in{\mathbb{S}}^{n}\colon x_{1}=0\}\cong{\mathbb{S}}^{n-1}$, and $M_{\pm}=\{x\in{\mathbb{S}}^{n}\colon\pm x_{1}>0\}$ are the two connected components of $M\setminus M^{\sigma}$.

Now replace $M$ by the projective space

[TABLE]

Using projective coordinates we have

[TABLE]

and $({\mathbb{P}}^{n})^{\sigma}\cong{\mathbb{P}}^{n-1}$, the complement of the fixed point set is the open cell

[TABLE]

hence connected.

Definition 2.7**.**

Let ${\mathfrak{g}}$ be a Lie algebra and $\tau\mathrel{\mathop{\ordinarycolon}}{\mathfrak{g}}\to{\mathfrak{g}}$ an involutive automorphism. We write ${\mathfrak{h}}=\ker(\tau-\mathbf{1})$ and ${\mathfrak{q}}=\ker(\tau+1)$, so that ${\mathfrak{g}}={\mathfrak{h}}+{\mathfrak{q}}$.

(a) The pair $({\mathfrak{g}},\tau)$ as well as the pair $({\mathfrak{g}},{\mathfrak{h}})$, is called a symmetric pair.

(b) The Cartan dual of $({\mathfrak{g}},\tau)$ is the symmetric pair $({\mathfrak{g}}^{c},\tau^{c})$ given by ${\mathfrak{g}}^{c}\mathrel{\mathop{\ordinarycolon}}={\mathfrak{h}}+i{\mathfrak{q}}$ and $\tau^{c}(x+iy)=x-iy$ for $x\in{\mathfrak{h}}$ and $y\in{\mathfrak{q}}$.

(c) If $({\mathfrak{g}},\tau,\sigma)$ a triple, where $\tau$ and $\sigma\mathrel{\mathop{\ordinarycolon}}{\mathfrak{g}}\to{\mathfrak{g}}$ are commuting involutive automorphisms, then the Cartan dual of $({\mathfrak{g}},\tau,\sigma)$ is the triple $({\mathfrak{g}}^{c},\tau^{c},\sigma^{c})$ with $\tau^{c}(x+iy)=x-iy$ and $\sigma^{c}(x+iy)=x-iy$ for $x\in{\mathfrak{h}},y\in{\mathfrak{q}}$.

Remark 2.8**.**

(a) If $({\mathfrak{g}},\tau,\sigma)$ is an irreducible dissecting triple and $\tau=\sigma$, then (2.2) implies $\mathop{{\rm dim}}\nolimits{\mathfrak{g}}^{-\tau}=1$, so that ${\mathfrak{g}}^{-\tau}$ is a one-dimensional ideal of ${\mathfrak{g}}$. Hence $\tau\not=\sigma$ if $({\mathfrak{g}},\tau,\sigma)$ is semisimple, irreducible and dissecting.

(b) If $({\mathfrak{g}},\tau)$ is irreducible and ${\mathfrak{g}}$ is not semisimple, then $\mathop{{\rm dim}}\nolimits{\mathfrak{g}}=1$ and $\tau=-\mathop{{\rm id}}\nolimits_{\mathfrak{g}}$. In fact, the radical ${\mathfrak{r}}\mathrel{\mathop{\ordinarycolon}}=\mathop{{\rm rad}}\nolimits({\mathfrak{g}})$ is a non-zero $\tau$-invariant ideal, hence equal to ${\mathfrak{g}}$. As its commutator algebra $[{\mathfrak{g}},{\mathfrak{g}}]$ is proper, it is zero, so that ${\mathfrak{g}}$ is abelian. As every one-dimensional subspace of ${\mathfrak{g}}^{-\tau}\not=\{0\}$ is an ideal, ${\mathfrak{g}}={\mathfrak{g}}^{-\tau}$ is one-dimensional.

(c) Clearly, $({\mathfrak{g}},\tau,\sigma)$ is dissecting if and only if $({\mathfrak{g}},\sigma,\tau)$ is dissecting. It is easy to see that $({\mathfrak{g}},\tau,\sigma)$ is dissecting if and only if the Cartan dual ($c$-dual) $({\mathfrak{g}}^{c},\tau^{c},\sigma^{c})$ is dissecting because $({\mathfrak{g}}^{c})^{-\tau^{c}}\cap({\mathfrak{g}}^{c})^{-\sigma^{c}}=i({\mathfrak{g}}^{-\tau}\cap{\mathfrak{g}}^{-\sigma})$.

(d) The dual symmetric Lie algebra $({\mathfrak{g}}^{c},\tau^{c})$ is irreducible if and only if $({\mathfrak{g}},\tau)$ is. In fact, $\tau$-invariant ideals of ${\mathfrak{g}}$ are of the form ${\mathfrak{j}}={\mathfrak{j}}_{\mathfrak{h}}\oplus{\mathfrak{j}}_{\mathfrak{q}}$, where

[TABLE]

and these conditions are equivalent to ${\mathfrak{j}}^{c}\mathrel{\mathop{\ordinarycolon}}={\mathfrak{j}}_{\mathfrak{h}}\oplus i\cdot{\mathfrak{j}}_{\mathfrak{q}}$ being an ideal of ${\mathfrak{g}}^{c}$.

3. Examples of

symmetric spaces with dissecting involutions

Let $(V,\beta)$ be a finite dimensional real vector space, endowed with a non-degenerate symmetric bilinear form $\beta$. Then every anisotropic element $x\in V$ defines an involution

[TABLE]

fixing $x^{\bot}=\{y\in V\colon\beta(x,y)=0\}$ pointwise, and for which ${\mathbb{R}}x$ is the $(-1)$-eigenspace.

Example 3.1**.**

If we consider $(V,\beta)$ as a flat semi-Riemannian symmetric space with respect to the point reflections $s_{z}(y)=2z-y$, then each $\sigma_{x}$ defines a dissecting isometric involution on $(V,\beta)$.

Example 3.2**.**

For $c\in{\mathbb{R}}^{\times}$, let

[TABLE]

denote the corresponding quadric in $(V,\beta)$. Then $(Q_{c},(-\sigma_{x})_{x\in Q_{c}})$ is a symmetric space (Definition 2.3) and $\mathop{{\rm dim}}\nolimits Q_{c}=\mathop{{\rm dim}}\nolimits V-1$.

If $\mathop{{\rm dim}}\nolimits V=1$, then $Q_{c}$ is discrete and contains at most two points, so let us assume that $\mathop{{\rm dim}}\nolimits V>1$. Let $M\subseteq Q_{c}$ be a connected component and let $x\in M$. Then $x^{\bot}$ contains a anisotropic element $y$ and the relations $\sigma_{y}(x)=x$ and $\sigma_{y}(Q_{c})=Q_{c}$ show that $\sigma_{y}|_{M}$ defines an involutive automorphism of the symmetric space $M$ for which $M^{\sigma_{y}}=M\cap y^{\bot}$ is non-empty. Moreover, $T_{x}(M\cap y^{\bot})=T_{x}(M)\cap y^{\bot}=\{x,y\}^{\bot}$ is a hyperplane in $T_{x}(M)=x^{\bot}$. Therefore $M^{\sigma_{y}}$ is a hypersurface whose complement contains the two open subsets

[TABLE]

satisfying $\sigma_{y}(M_{\pm})=M_{\mp}$. Thus $\sigma_{y}$ defines a dissecting involution of $M$ (Definition 2.1). That $M_{\pm}$ are in fact connected follows from a quick inspection of the possible quadrics in $3$-dimensional spaces.

The space $(V,\beta)$ is isometric to some $({\mathbb{R}}^{p+q},\beta_{p,q})$ with

[TABLE]

We can always assume that $c=1$ because dilation by $r\in{\mathbb{R}}^{\times}$ is an isomorphism of symmetric spaces from $Q_{c}$ to $Q_{r^{2}c}$ and the quadric $Q_{-1}$ in $({\mathbb{R}}^{p+q},\beta_{p,q})$ is isomorphic to the quadric $Q_{1}$ in $({\mathbb{R}}^{p+q},\beta_{q,p})$. In particular, all these quadrics have isomorphic symmetry groups $\mathop{\rm O{}}\nolimits_{p,q}({\mathbb{R}})\cong\mathop{\rm O{}}\nolimits_{q,p}({\mathbb{R}})$. Thus, up to isomorphisms of symmetric spaces, we obtain the connected symmetric spaces $M^{p,q}\cong\mathop{{\rm SO}}\nolimits_{p,q}({\mathbb{R}})_{0}/\mathop{{\rm SO}}\nolimits_{p-1,q}({\mathbb{R}})_{0}$ as connected components of the quadric $Q_{1}$. According to the sign of $\beta(y,y)$, we obtain two types of dissecting involutions on $M^{p,q}$. This leads to two types of symmetric spaces with dissecting involutions.

Let $n\mathrel{\mathop{\ordinarycolon}}=p+q-1=\mathop{{\rm dim}}\nolimits M\geq 1$.

• •

For $q=0$, we obtain the sphere $M^{n+1,0}={\mathbb{S}}^{n}\cong\mathop{{\rm SO}}\nolimits_{n+1}({\mathbb{R}})/\mathop{{\rm SO}}\nolimits_{n}({\mathbb{R}})$ with an isometric reflection $\sigma_{e_{1}}$ with fixed point set ${\mathbb{S}}^{n-1}$ (an equator).

• •

For $p=1$, we obtain the hyperbolic space $M^{1,n}={\mathbb{H}}^{n}\cong\mathop{{\rm SO}}\nolimits_{1,n}({\mathbb{R}})/\mathop{{\rm SO}}\nolimits_{n}({\mathbb{R}})$ with an isometric reflection with fixed point set ${\mathbb{H}}^{n-1}$.

• •

For $q=1$, we obtain de Sitter space $M^{n,1}\cong\mathop{{\rm dS}}\nolimits^{n}\cong\mathop{{\rm SO}}\nolimits_{n,1}({\mathbb{R}})/\mathop{{\rm SO}}\nolimits_{n-1,1}({\mathbb{R}})$ and two possible fixed point sets, one isomorphic to $M^{n,0}={\mathbb{S}}^{n-1}\cong\mathop{{\rm SO}}\nolimits_{n}({\mathbb{R}})/\mathop{{\rm SO}}\nolimits_{n-1}({\mathbb{R}})$ and the other to $M^{n-1,1}=\mathop{{\rm dS}}\nolimits^{n-1}\cong\mathop{{\rm SO}}\nolimits_{n-1,1}({\mathbb{R}})/\mathop{{\rm SO}}\nolimits_{n-2,1}({\mathbb{R}})$. They correspond to the reflections $\sigma_{y}$ with $y=e_{n+1}$ and $e_{n}$, respectively.

For $p,q>1$ and the quadric $M^{p,q}$, we obtain fixed point manifolds of type $M^{p-1,q}$ and $M^{p,q-1}$. Note that $M^{p,q}$ is diffeomorphic to ${\mathbb{S}}^{p-1}\times{\mathbb{R}}^{q}$, hence simply connected if $p\geq 3$.

For the Lie algebra ${\mathfrak{g}}=\mathop{{\mathfrak{so}}}\nolimits_{p,q}({\mathbb{R}})$ and an involution $\tau$ induced by a reflection $\sigma_{e_{j}}$ in $e_{j}^{\bot}$ for a basis vector $e_{j}$, we have

[TABLE]

This follows easily by realizing ${\mathfrak{g}}^{c}$ on the subspace ${\mathbb{R}}ie_{j}\oplus\sum_{k\not=j}{\mathbb{R}}e_{k}\subseteq{\mathbb{C}}^{p+q}$. Therefore the class of dissecting triples obtained this way is stable under duality.

For $G=\mathop{{\rm SO}}\nolimits_{p,q}({\mathbb{R}})_{0}$, we thus obtain dissecting triples $({\mathfrak{g}},\tau,\sigma)$, where $\tau$ and $\sigma$ are induced by reflections in hyperplanes orthogonal to coordinate axes. All these examples are irreducible and we shall see that all irreducible semisimple dissecting triples are of this type.

Example 3.3**.**

On the space $V\mathrel{\mathop{\ordinarycolon}}=M_{2}({\mathbb{R}})$ of real $(2\times 2)$-matrices, the determinant defines a quadratic form. From the relation $X^{2}-(\mathop{{\rm tr}}\nolimits X)X+\det(X)\mathbf{1}=0$, we derive that $\det(X)=\beta(X,X)$ for

[TABLE]

This form is positive definite on ${\mathbb{R}}\mathbf{1}\oplus\mathop{{\mathfrak{so}}}\nolimits_{2}({\mathbb{R}})$ and negative definite on its orthogonal complement $\mathop{{\mathfrak{sl}}}\nolimits_{2}({\mathbb{R}})\cap\mathop{{\rm Herm}}\nolimits_{2}({\mathbb{R}})$, hence of signature $(p,q)=(2,2)$. Therefore

[TABLE]

is a $3$-dimensional quadric and the $6$-dimensional group $G\mathrel{\mathop{\ordinarycolon}}=\mathop{{\rm SO}}\nolimits_{2,2}({\mathbb{R}})_{0}$ acts transitively on $M$. It contains the subgroup $\mathop{{\rm SL}}\nolimits_{2}({\mathbb{R}})_{L}$ acting by left multiplications, and the subgroup $\mathop{{\rm SL}}\nolimits_{2}({\mathbb{R}})_{R}$, acting by right multiplications. Clearly, $\mathop{{\rm SL}}\nolimits_{2}({\mathbb{R}})_{L}\cap\mathop{{\rm SL}}\nolimits_{2}({\mathbb{R}})_{R}=\{\pm\mathbf{1}\}$, so that $G=\mathop{{\rm SL}}\nolimits_{2}({\mathbb{R}})_{L}\mathop{{\rm SL}}\nolimits_{2}({\mathbb{R}})_{R}$ for dimensional reasons.

Let $x\mathrel{\mathop{\ordinarycolon}}=\mathbf{1}$, $y_{1}\mathrel{\mathop{\ordinarycolon}}=\begin{pmatrix}0&-1\\ 1&0\end{pmatrix}$ and $y_{2}\mathrel{\mathop{\ordinarycolon}}=\begin{pmatrix}1&0\\ 0&-1\end{pmatrix}$. Then $\beta(y_{1},y_{1})=1$, $\beta(y_{2},y_{2})=-1$, and $y_{1/2}$ are both orthogonal to $x=\mathbf{1}$. We thus obtain two dissecting involutions $\sigma^{M}_{1/2}$ on $M=\mathop{{\rm SL}}\nolimits_{2}({\mathbb{R}})$ from the corresponding orthogonal reflections. Their action on the Lie algebra $\mathop{{\mathfrak{sl}}}\nolimits_{2}({\mathbb{R}})\cong T_{\mathbf{1}}(M)$ is given by

[TABLE]

The corresponding involutions on the symmetric space $M=\mathop{{\rm SL}}\nolimits_{2}({\mathbb{R}})$ are

[TABLE]

As $\sigma_{1}^{G}(g_{1}gg_{2}^{-1})=g_{2}^{-\top}\sigma_{1}^{G}(g)g_{1}^{\top}$ and $\sigma_{2}^{G}(g_{1}gg_{2}^{-1})=I_{1,1}g_{2}I_{1,1}\sigma_{2}^{G}(g)I_{1,1}g_{1}^{-1}I_{1,1}$, the involutions induced on the Lie algebra ${\mathfrak{g}}\cong\mathop{{\mathfrak{sl}}}\nolimits_{2}({\mathbb{R}})\oplus\mathop{{\mathfrak{sl}}}\nolimits_{2}({\mathbb{R}})$ are given by

[TABLE]

Example 3.4**.**

On the four-dimensional space $V\mathrel{\mathop{\ordinarycolon}}={\mathbb{H}}={\mathbb{R}}\mathbf{1}\oplus\mathop{{\mathfrak{su}}}\nolimits_{2}({\mathbb{C}})\subseteq M_{2}({\mathbb{C}})$ of quaternions, we consider the scalar product given by

[TABLE]

Then

[TABLE]

is a $3$-dimensional sphere.

The $6$-dimensional group $G\mathrel{\mathop{\ordinarycolon}}=\mathop{{\rm SO}}\nolimits_{4}({\mathbb{R}})_{0}$ acts transitively on $M$. It contains the subgroup $\mathop{{\rm SU}}\nolimits_{2}({\mathbb{C}})_{L}$ acting by left multiplications, and the subgroup $\mathop{{\rm SU}}\nolimits_{2}({\mathbb{C}})_{R}$, acting by right multiplications. As above, $\mathop{{\rm SU}}\nolimits_{2}({\mathbb{C}})_{L}\cap\mathop{{\rm SU}}\nolimits_{2}({\mathbb{C}})_{R}=\{\pm\mathbf{1}\}$ and $G=\mathop{{\rm SU}}\nolimits_{2}({\mathbb{C}})_{L}\mathop{{\rm SU}}\nolimits_{2}({\mathbb{C}})_{R}$. As $x\mathrel{\mathop{\ordinarycolon}}=\mathbf{1}$ and $y\mathrel{\mathop{\ordinarycolon}}=\begin{pmatrix}i&0\\ 0&-i\end{pmatrix}$ are orthogonal unit vectors, we obtain a dissecting involution $\sigma^{M}$ on $M=\mathop{{\rm SU}}\nolimits_{2}({\mathbb{C}})$ from the orthogonal reflections $\sigma_{y}$. Its action on the Lie algebra $\mathop{{\mathfrak{su}}}\nolimits_{2}({\mathbb{C}})\cong T_{\mathbf{1}}(M)$ is given by

[TABLE]

The corresponding involution on the symmetric space $M=\mathop{{\rm SU}}\nolimits_{2}({\mathbb{C}})$ is $\sigma^{G}(g)=I_{1,1}g^{-1}I_{1,1},$ and the involution induced on the Lie algebra ${\mathfrak{g}}\cong\mathop{{\mathfrak{su}}}\nolimits_{2}({\mathbb{C}})\oplus\mathop{{\mathfrak{su}}}\nolimits_{2}({\mathbb{C}})$ is given by

[TABLE]

4. The classification

In this subsection we show that all irreducible dissecting triples $({\mathfrak{g}},\tau,\sigma)$ come from the quadrics as in Example 3.2. Our strategy is to first reduce the classification to the case where ${\mathfrak{g}}$ is simple. Then we reduce it to the case where $({\mathfrak{g}},\tau)$ is Riemannian and eventually we have to classify certain triples of commuting involutions on $\mathop{{\mathfrak{so}}}\nolimits_{n}({\mathbb{R}})$. Here Riemannian means that there exists a Riemannian symmetric space $M=G/H$ such that the Lie algebra of $G$ is ${\mathfrak{g}}$ and the Lie algebra of $H$ is ${\mathfrak{h}}$.

We start with some simple observations concerning commuting involutions $\sigma$ and $\tau$ on a Lie algebra ${\mathfrak{g}}$. Let ${\mathfrak{h}}={\mathfrak{g}}^{\tau}$, ${\mathfrak{q}}={\mathfrak{g}}^{-\tau}$, ${\mathfrak{l}}={\mathfrak{g}}^{\sigma}$, and ${\mathfrak{m}}={\mathfrak{g}}^{-\sigma}$. Then ${\mathfrak{g}}$ decomposes as

[TABLE]

where the indices ${\mathfrak{l}}$ and ${\mathfrak{m}}$, resp., indicate intersections with ${\mathfrak{l}}$ and ${\mathfrak{m}}$, respectively. We assume that the symmetric Lie algebra $({\mathfrak{g}},\tau)$ is irreducible and non-abelian (Remark 2.8(b)).

Definition 4.1**.**

If ${\mathfrak{g}}$ is semisimple, then $\sigma$ and $\tau$ generate a finite subgroup of $\mathop{{\rm Aut}}\nolimits({\mathfrak{g}})$, so that [HN12, Prop. 13.2.14] implies that there exists a Cartan involution $\theta$ of ${\mathfrak{g}}$, commuting with $\tau$ and $\sigma$. We write ${\mathfrak{g}}^{\theta}={\mathfrak{k}}$ and ${\mathfrak{g}}^{-\theta}={\mathfrak{p}}$ for the eigenspaces of $\theta$.

The compact Lie algebra ${\mathfrak{g}}^{r}\mathrel{\mathop{\ordinarycolon}}={\mathfrak{k}}\oplus i{\mathfrak{p}}$ is called the compact dual of ${\mathfrak{g}}$. It carries three commuting involutions $\theta^{r},\tau^{r}$ and $\sigma^{r}$, defined by extending each involution to a complex linear involution on ${\mathfrak{g}}_{\mathbb{C}}$ and then restrict to ${\mathfrak{g}}^{r}$.

Lemma 4.2**.**

The triple $({\mathfrak{g}}^{r},\tau^{r},\sigma^{r})$ is dissecting if and only if $({\mathfrak{g}},\tau,\sigma)$ is.

Proof.

First we observe that

[TABLE]

As ${\mathfrak{q}}_{\mathfrak{m}}$ is $\theta$-invariant, ${\mathfrak{q}}_{\mathfrak{m}}=({\mathfrak{q}}_{\mathfrak{m}}\cap{\mathfrak{k}})\oplus({\mathfrak{q}}_{\mathfrak{m}}\cap{\mathfrak{p}})$. Hence it is one-dimensional if and only if $({\mathfrak{g}}^{r})^{-\tau^{r}}\cap({\mathfrak{g}}^{r})^{-\sigma^{r}}$ is one-dimensional. ∎

We have the following possibilities for an irreducible non-abelian symmetric pair $({\mathfrak{g}},\tau)$ (see [F-J86, p. 6] for a discussion and [F-J86, pp. 9–11] for two commuting involutions):

• (1)

${\mathfrak{g}}$ is simple but not complex.111 We say that a real simple Lie algebra ${\mathfrak{g}}$ is complex if there exists a complex structure $I$ on ${\mathfrak{g}}$ commuting with $\mathop{{\rm ad}}\nolimits{\mathfrak{g}}$, so that $ix\mathrel{\mathop{\ordinarycolon}}=Ix$ turns ${\mathfrak{g}}$ into a complex Lie algebra. Recall that all complex simple Lie algebras are also simple as real Lie algebras.

• (2)

${\mathfrak{g}}$ is a simple complex Lie algebra and $\tau$ is complex linear.

• (3)

${\mathfrak{g}}$ is a simple complex Lie algebra and $\tau$ is antilinear, i.e., a conjugation with respect to the real form ${\mathfrak{g}}_{1}={\mathfrak{g}}^{\tau}$, ${\mathfrak{g}}\cong{\mathfrak{g}}_{1}\oplus i{\mathfrak{g}}_{1}$ and $\tau(x+iy)=x-iy$ for $x,y\in{\mathfrak{g}}_{1}$.

• (4)

${\mathfrak{g}}={\mathfrak{g}}_{1}\oplus{\mathfrak{g}}_{1}$ with ${\mathfrak{g}}_{1}$ simple and $\tau(x,y)=(y,x)$.

In fact, if ${\mathfrak{g}}$ is not simple, then $\tau$ permutes the simple ideals of ${\mathfrak{g}}$ in a non-trivial way, and since $\tau$ is an involution, irreducibility implies that ${\mathfrak{g}}={\mathfrak{g}}_{1}\oplus{\mathfrak{g}}_{2}$ for two simple ideals satisfying $\tau({\mathfrak{g}}_{1})={\mathfrak{g}}_{2}$. Then ${\mathfrak{g}}_{2}\cong{\mathfrak{g}}_{1}$, and $({\mathfrak{g}},\tau)$ takes the form (4).

Lemma 4.3**.**

The pairs $({\mathfrak{g}},\tau)$ in (3), and the pairs $({\mathfrak{g}}={\mathfrak{g}}_{1}\oplus{\mathfrak{g}}_{1},\tau)$ as in (4), where ${\mathfrak{g}}_{1}$ is simple but not complex, are $c$-dual to each other.

Proof.

If $({\mathfrak{g}},\tau)$ is of type (4), then ${\mathfrak{g}}_{\mathbb{C}}\cong{\mathfrak{g}}_{1,{\mathbb{C}}}\oplus{\mathfrak{g}}_{1,{\mathbb{C}}}$. For $z=x+iy\in{\mathfrak{g}}_{1,{\mathbb{C}}}$ we write $\overline{z}\mathrel{\mathop{\ordinarycolon}}=x-iy$. Then ${\mathfrak{g}}^{c}=\{(z,\overline{z})\colon z\in{\mathfrak{g}}_{1,{\mathbb{C}}}\}$, and ${\mathfrak{g}}_{1,{\mathbb{C}}}\to{\mathfrak{g}}^{c},z\mapsto(z,\overline{z})$ is an isomorphism of real Lie algebras. The Lie algebra ${\mathfrak{g}}^{c}$ is simple if and only if ${\mathfrak{g}}_{1}$ is not complex.

If, conversely, $({\mathfrak{g}},\tau)=({\mathfrak{g}}_{1,{\mathbb{C}}},\tau)$, where $\tau$ is complex conjugation with respect to ${\mathfrak{g}}_{1}$ as in (3), then the inclusion ${\mathfrak{g}}={\mathfrak{g}}_{1,{\mathbb{C}}}\hookrightarrow{\mathfrak{g}}_{1,{\mathbb{C}}}\oplus{\mathfrak{g}}_{1,{\mathbb{C}}},z\mapsto(z,\tau(z))$ extends to an isomorphism of complex Lie algebras ${\mathfrak{g}}_{\mathbb{C}}\to{\mathfrak{g}}_{1,{\mathbb{C}}}\oplus{\mathfrak{g}}_{1,{\mathbb{C}}}$. Accordingly,

[TABLE]

Since ${\mathfrak{g}}\cong{\mathfrak{g}}_{1,{\mathbb{C}}}$ is simple, ${\mathfrak{g}}_{1}$ is not complex. ∎

Lemma 4.4**.**

Let ${\mathfrak{g}}_{1}$ be a simple real Lie algebra and consider on ${\mathfrak{g}}\mathrel{\mathop{\ordinarycolon}}={\mathfrak{g}}_{1}\oplus{\mathfrak{g}}_{1}$ the flip involution $\tau(x,y)=(y,x)$. Then, for any involution $\sigma$ of ${\mathfrak{g}}$ commuting with $\tau$, there exists an involutive automorphism $\sigma_{1}$ of ${\mathfrak{g}}_{1}$ such that either

[TABLE]

Proof.

As $\tau$ and $\sigma$ commute, $\sigma$ leaves the diagonal $\{(x,x)\colon x\in{\mathfrak{g}}_{1}\}={\mathfrak{g}}^{\tau}$ invariant. Hence there exists an involutive automorphism $\sigma_{1}\in\mathop{{\rm Aut}}\nolimits({\mathfrak{g}}_{1})$ with

[TABLE]

Now $\widetilde{\sigma}(x,y)=(\sigma_{1}x,\sigma_{1}y)$ defines an involutive automorphism of ${\mathfrak{g}}$ commuting with $\tau$. We show that either $\sigma=\widetilde{\sigma}$ or $\sigma=\tau\widetilde{\sigma}$.

We consider the automorphism $\gamma\mathrel{\mathop{\ordinarycolon}}=\sigma\widetilde{\sigma}\in\mathop{{\rm Aut}}\nolimits({\mathfrak{g}})$ which commutes with $\tau$ and fixes the diagonal ${\mathfrak{g}}^{\tau}$ pointwise. Hence $\gamma$ preserves the anti-diagonal and there exists a linear map $c\colon{\mathfrak{g}}_{1}\to{\mathfrak{g}}_{1}$ with

[TABLE]

As $\gamma$ commutes with $\mathop{{\rm ad}}\nolimits(x,x)$ for $x\in{\mathfrak{g}}_{1}$, it follows that

[TABLE]

We also have

[TABLE]

so that

[TABLE]

As ${\mathfrak{g}}$ is assumed to be simple, and hence $[{\mathfrak{g}}_{1},{\mathfrak{g}}_{1}]={\mathfrak{g}}_{1}$, it follows that $c^{2}=\mathop{{\rm id}}\nolimits_{{\mathfrak{g}}_{1}}$.

Since $c$ commutes with $\mathop{{\rm ad}}\nolimits{\mathfrak{g}}_{1}$, the eigenspaces $\ker(c\pm\mathbf{1})$ are ideals of the simple Lie algebra ${\mathfrak{g}}_{1}$, so that $c\in\{\pm\mathop{{\rm id}}\nolimits_{{\mathfrak{g}}_{1}}\}$. We conclude that either $c=1$, i.e., $\sigma=\widetilde{\sigma}$, or $c=-1$, which means that $\sigma=\tau\widetilde{\sigma}$. ∎

The triple $({\mathfrak{g}},\tau,\sigma)$ is dissecting if and only if $\mathop{{\rm dim}}\nolimits{\mathfrak{q}}_{\mathfrak{m}}=1$. Let $0\not=x_{0}\in{\mathfrak{q}}_{\mathfrak{m}}$. Recall that $x\in{\mathfrak{g}}$ is said to be elliptic if $\mathop{{\rm ad}}\nolimits x$ is semisimple with purely imaginary eigenvalues, and hyperbolic if $\mathop{{\rm ad}}\nolimits x$ is diagonalizable over ${\mathbb{R}}$.

Lemma 4.5**.**

For an irreducible non-abelian dissecting triple $({\mathfrak{g}},\tau,\sigma)$, we have:

• (i)

${\mathfrak{g}}^{\sigma\tau}={\mathfrak{h}}_{\mathfrak{l}}\oplus{\mathbb{R}}x_{0}$, ${\mathfrak{h}}_{\mathfrak{l}}$ is an ideal of ${\mathfrak{g}}^{\sigma\tau}$ and $x_{0}$ is central in ${\mathfrak{g}}^{\sigma\tau}$.

• (ii)

$x_{0}$* is either elliptic or hyperbolic.*

Note that, if $x_{0}\in{\mathfrak{q}}$ is elliptic, then $ix_{0}\in{\mathfrak{g}}^{c}={\mathfrak{h}}\oplus i{\mathfrak{q}}$ is hyperbolic.

Proof.

(i) We have ${\mathfrak{g}}^{\sigma\tau}={\mathfrak{h}}_{\mathfrak{l}}\oplus{\mathfrak{q}}_{\mathfrak{m}}={\mathfrak{h}}_{\mathfrak{l}}\oplus{\mathbb{R}}x_{0}$. Hence $[{\mathfrak{g}}^{\sigma\tau},{\mathbb{R}}x_{0}]=[{\mathfrak{h}}_{\mathfrak{l}},{\mathfrak{q}}_{\mathfrak{m}}]\subseteq{\mathfrak{q}}_{\mathfrak{m}}={\mathbb{R}}x_{0},$ so that ${\mathbb{R}}x_{0}$ is an ideal in ${\mathfrak{g}}^{\sigma\tau}$. As ${\mathfrak{g}}^{\sigma\tau}$ is reductive,222For any involution $\tau$ of a semisimple Lie algebra ${\mathfrak{g}}$, the fixed point algebra is reductive in ${\mathfrak{g}}$. Since there exists a Cartan involution $\theta$ commuting with $\tau$, this follows from [Wa72, Cor. 1.1.5.4]. all its one-dimensional ideals are central. Therefore $x_{0}$ is central and thus ${\mathfrak{h}}_{\mathfrak{l}}$ also is an ideal.

(ii) According to [Wa72, Prop. 1.3.5.1], $x_{0}$ has a Jordan decomposition $x_{0}=x_{s}+x_{n}$, where $x_{s}\in{\mathfrak{g}}$ is semisimple and $x_{n}\in{\mathfrak{g}}$ is nilpotent. Applying $\sigma$, the uniqueness of the Jordan decomposition and

[TABLE]

imply that $\sigma(x_{s})=-x_{s}$ and $\sigma(x_{n})=-x_{n}$. We likewise obtain $\tau(x_{s})=-x_{s}$ and $\tau(x_{n})=-x_{n}$. As $\mathop{{\rm dim}}\nolimits{\mathfrak{q}}_{\mathfrak{m}}=1$, we have $x_{0}=x_{s}$ or $x_{0}=x_{n}$.

As ${\mathfrak{g}}^{\sigma\tau}$ is reductive in ${\mathfrak{g}}$, the Lie algebra decomposes into simple ${\mathfrak{g}}^{\sigma\tau}$-submodules. If the central element $x_{0}\in{\mathfrak{g}}^{\sigma\tau}$ is nilpotent, it acts trivially on all these simple submodules, so that $\mathop{{\rm ad}}\nolimits x_{0}=0$. As ${\mathfrak{g}}$ is semisimple and $x_{0}\not=0$, this cannot happen, and we conclude that $x_{0}$ is semisimple.

Now we decompose $x_{0}=x_{e}+x_{h}$ where $x_{e}$ and $x_{h}$ commute, $x_{e}$ is elliptic and $x_{h}$ is hyperbolic. The uniqueness of the decomposition implies as above that $x_{e},x_{h}\in{\mathfrak{q}}_{\mathfrak{m}}$, hence that $x_{0}$ is either elliptic or hyperbolic. ∎

Let us now reduce the classification to the case where ${\mathfrak{g}}$ is simple. This is achieved by the following lemma, which reduces the case where ${\mathfrak{g}}$ is not simple to $\mathop{{\mathfrak{su}}}\nolimits_{2}({\mathbb{C}})\cong\mathop{{\mathfrak{so}}}\nolimits_{3}({\mathbb{R}})$ and $\mathop{{\mathfrak{sl}}}\nolimits_{2}({\mathbb{R}})\cong\mathop{{\mathfrak{so}}}\nolimits_{1,2}({\mathbb{R}})$ which is discussed in Examples 3.3 and 3.4. We first take a closer look at these two Lie algebras.

Remark 4.6**.**

The involutions on $\mathop{{\mathfrak{su}}}\nolimits_{2}({\mathbb{C}})$ and $\mathop{{\mathfrak{sl}}}\nolimits_{2}({\mathbb{R}})$ are determined by their one-dimensional fixed point algebras ${\mathfrak{g}}_{1}^{\sigma_{1}}={\mathbb{R}}z_{0}$, where $z_{0}$ can be either elliptic of hyperbolic for $\mathop{{\mathfrak{sl}}}\nolimits_{2}({\mathbb{R}})$, and $z_{0}$ is elliptic for $\mathop{{\mathfrak{su}}}\nolimits_{2}({\mathbb{C}})\cong\mathop{{\mathfrak{so}}}\nolimits_{3}({\mathbb{R}})$. As $\mathop{{\rm Aut}}\nolimits(\mathop{{\mathfrak{su}}}\nolimits_{2}({\mathbb{C}}))=\mathop{{\rm Aut}}\nolimits(\mathop{{\mathfrak{so}}}\nolimits_{3}({\mathbb{R}}))\cong\mathop{{\rm SO}}\nolimits_{3}({\mathbb{R}})$, involutive automorphisms correspond to non-trivial involutions $\sigma\in\mathop{{\rm SO}}\nolimits_{3}({\mathbb{R}})$. As $\det\sigma=1$, these reflections have $1$-dimensional fixed point spaces, hence are mutually conjugate, so that there is only one equivalence class for $\mathop{{\mathfrak{su}}}\nolimits_{2}({\mathbb{C}})$. We also observe that complex conjugation on $\mathop{{\mathfrak{su}}}\nolimits_{2}({\mathbb{C}})$ coincides with $x\mapsto-x^{\top}$ and with conjugation by $\begin{pmatrix}0&1\\ -1&0\end{pmatrix}$.

Lemma 4.7**.**

Assume that $({\mathfrak{g}},\tau,\sigma)$ is an irreducible non-abelian dissecting triple. If ${\mathfrak{g}}$ is not simple, then $({\mathfrak{g}},\tau,\sigma)$ is equivalent to

[TABLE]

where either

• (a)

${\mathfrak{g}}_{1}=\mathop{{\mathfrak{su}}}\nolimits_{2}({\mathbb{C}})\simeq\mathop{{\mathfrak{so}}}\nolimits_{3}({\mathbb{R}})$* and $\sigma_{1}(x)=\overline{x}$.*

• (b)

${\mathfrak{g}}_{1}=\mathop{{\mathfrak{sl}}}\nolimits_{2}({\mathbb{R}})\simeq\mathop{{\mathfrak{so}}}\nolimits_{2,1}({\mathbb{R}})$* and $\sigma_{1}(x)=-x^{\top}$ or $\sigma_{1}(x)=\mathrm{I}_{1,1}x\mathrm{I}_{1,1}.$*

Proof.

Since ${\mathfrak{g}}$ is not simple, $({\mathfrak{g}},\tau)$ is equivalent to a pair of the form ${\mathfrak{g}}={\mathfrak{g}}_{1}\oplus{\mathfrak{g}}_{1}$ with ${\mathfrak{g}}_{1}$ simple and $\tau(x,y)=(y,x)$. Lemma 4.4 shows that there exists an involutive automorphism $\sigma_{1}$ of ${\mathfrak{g}}_{1}$ such that either

[TABLE]

Let $\widetilde{\sigma}(x,y)=(\sigma_{1}(x),\sigma_{1}(y))$ be as in the proof of Lemma 4.4. We then have

[TABLE]

Write $x_{0}=(y_{0},-y_{0})$. In the first case ${\mathfrak{g}}_{1}^{-\sigma_{1}}={\mathbb{R}}y_{0}$ is one dimensional. Hence ${\mathfrak{g}}_{1}={\mathfrak{g}}_{1}^{\sigma_{1}}\oplus{\mathbb{R}}y_{0}$ and

[TABLE]

It follows that ${\mathbb{R}}y_{0}$ is an ideal, contradicting the assumption that ${\mathfrak{g}}_{1}$ is simple. Hence we are in the second case where $\sigma=\tau\widetilde{\sigma}$ and ${\mathfrak{q}}_{\mathfrak{m}}\cong{\mathfrak{g}}_{1}^{\sigma_{1}}={\mathbb{R}}y_{0}$ is one-dimensional. It therefore remains to determine the symmetric pairs $({\mathfrak{g}}_{1},\sigma_{1})$, where ${\mathfrak{g}}_{1}$ is simple and ${\mathfrak{g}}_{1}^{\sigma_{1}}$ is one-dimensional.

Case 1: ${\mathfrak{g}}_{1}^{-\sigma_{1}}$ is an irreducible module of ${\mathbb{R}}y_{0}={\mathfrak{g}}_{1}^{\sigma_{1}}$. Then $\mathop{{\rm dim}}\nolimits{\mathfrak{g}}_{1}^{-\sigma_{1}}\leq 2$, and since $\mathop{{\rm dim}}\nolimits{\mathfrak{g}}_{1}\geq 3$, it follows that ${\mathfrak{g}}_{1}$ is $3$-dimensional, hence isomorphic to $\mathop{{\mathfrak{su}}}\nolimits_{2}({\mathbb{C}})$ or $\mathop{{\mathfrak{sl}}}\nolimits_{2}({\mathbb{R}})$. Up to equivalence, there are only two equivalence classes of three-dimensional symmetric Lie algebras $({\mathfrak{g}}_{1},\tau_{1})$ for which $\mathop{{\rm ad}}\nolimits y_{0}$ is irreducible on ${\mathfrak{g}}_{1}^{-\sigma_{1}}$, one for ${\mathfrak{g}}_{1}\cong\mathop{{\mathfrak{su}}}\nolimits_{2}({\mathbb{C}})$ and one for ${\mathfrak{g}}_{1}\cong\mathop{{\mathfrak{sl}}}\nolimits_{2}({\mathbb{R}})$ (Remark 4.6). Here $y_{0}$ is elliptic.

Case 2: ${\mathfrak{g}}_{1}^{-\sigma_{1}}$ is not irreducible under $\mathop{{\rm ad}}\nolimits y_{0}$. By [HÓ97, Lem. 1.3.4, p. 13], ${\mathfrak{g}}_{1}$ is $3$-graded, ${\mathfrak{g}}_{1}=V_{-1}\oplus V_{0}\oplus V_{1}$, where $V_{0}={\mathfrak{g}}_{1}^{\sigma_{1}}={\mathbb{R}}y_{0}$ and $V_{\pm 1}$ are irreducible ${\mathfrak{g}}^{\sigma_{1}}$-submodules of ${\mathfrak{g}}_{1}^{-\sigma_{1}}$ which are abelian Lie algebras. Then $Dv_{j}\mathrel{\mathop{\ordinarycolon}}=jv_{j}$ for $v_{j}\in V_{j}$ defines a derivation of ${\mathfrak{g}}_{1}$. As ${\mathfrak{g}}_{1}$ is semisimple, $D=\mathop{{\rm ad}}\nolimits h$ for some $h\in{\mathfrak{g}}_{1}$. Since $D$ commutes with $\sigma_{1}$, we have $\sigma_{1}(h)=h$. Therefore $h\in{\mathbb{R}}y_{0}$, and in particular $y_{0}$ is hyperbolic. Irreducibility of $V_{\pm 1}$ thus implies $\mathop{{\rm dim}}\nolimits V_{1}=\mathop{{\rm dim}}\nolimits V_{-1}=1$. Hence ${\mathfrak{g}}_{1}$ is a $3$-dimensional non-compact simple Lie algebra, so that ${\mathfrak{g}}_{1}\simeq\mathfrak{sl}_{2}({\mathbb{R}})$. ∎

We next discuss the case where ${\mathfrak{g}}$ is complex simple.

Corollary 4.8**.**

*If the triple $({\mathfrak{g}},\tau,\sigma)$ is dissecting and ${\mathfrak{g}}$ is complex simple, then ${\mathfrak{g}}\simeq\mathfrak{sl}_{2}({\mathbb{C}})\simeq\mathop{{\mathfrak{so}}}\nolimits_{3,1}({\mathbb{R}})$ and $\tau$ is the conjugation with respect to $\mathop{{\mathfrak{su}}}\nolimits_{2}({\mathbb{C}})\simeq\mathop{{\mathfrak{so}}}\nolimits_{3}({\mathbb{R}})$ or $\mathfrak{sl}_{2}({\mathbb{R}})\simeq\mathop{{\mathfrak{so}}}\nolimits_{2,1}({\mathbb{R}})$. In this case ${\mathfrak{g}}^{r}\cong\mathop{{\mathfrak{su}}}\nolimits_{2}({\mathbb{C}})\oplus\mathop{{\mathfrak{su}}}\nolimits_{2}({\mathbb{C}})\cong\mathop{{\mathfrak{so}}}\nolimits_{3}({\mathbb{R}})\oplus\mathop{{\mathfrak{so}}}\nolimits_{3}({\mathbb{R}})\cong\mathop{{\mathfrak{so}}}\nolimits_{4}({\mathbb{R}})$. *

Proof.

Since ${\mathfrak{g}}^{-\tau}\cap{\mathfrak{g}}^{-\sigma}$ is one-dimensional real, either $\tau$ or $\sigma$ is not complex linear. Since $({\mathfrak{g}},\sigma,\tau)$ is also dissecting, we may assume that $\tau$ is antilinear. Then ${\mathfrak{g}}_{1}\mathrel{\mathop{\ordinarycolon}}={\mathfrak{g}}^{\tau}$ is a real form of ${\mathfrak{g}}$ and ${\mathfrak{g}}\cong{\mathfrak{g}}_{1,{\mathbb{C}}}$. By Lemma 4.3, ${\mathfrak{g}}^{c}\mathrel{\mathop{\ordinarycolon}}={\mathfrak{g}}^{\tau}\oplus i{\mathfrak{g}}^{-\tau}\cong{\mathfrak{g}}_{1}\oplus{\mathfrak{g}}_{1}$ with $\tau^{c}(z,w)=(w,z)$, and $({\mathfrak{g}}^{c},\tau^{c},\sigma^{c})$ is dissecting by Remark 2.8(c). Further, $({\mathfrak{g}}^{c},\tau^{c})$ is irreducible (Remark 2.8(d)), so that Lemma 4.7 implies that ${\mathfrak{g}}\cong{\mathfrak{g}}_{1,{\mathbb{C}}}\cong\mathop{{\mathfrak{sl}}}\nolimits_{2}({\mathbb{C}})\cong\mathop{{\mathfrak{so}}}\nolimits_{1,3}({\mathbb{R}})$, where $\tau$ corresponds to the involution on $\mathop{{\mathfrak{so}}}\nolimits_{1,3}({\mathbb{R}})$ with fixed point algebra $\mathop{{\mathfrak{so}}}\nolimits_{3}({\mathbb{R}})$ or $\mathop{{\mathfrak{so}}}\nolimits_{1,2}({\mathbb{R}})$, the two real forms of ${\mathfrak{g}}$. The compact dual of ${\mathfrak{g}}=\mathop{{\mathfrak{sl}}}\nolimits_{2}({\mathbb{C}})$ is ${\mathfrak{g}}^{r}=\mathop{{\mathfrak{su}}}\nolimits_{2}({\mathbb{C}})\oplus\mathop{{\mathfrak{su}}}\nolimits_{2}({\mathbb{C}})\cong\mathop{{\mathfrak{so}}}\nolimits_{3}({\mathbb{R}})\oplus\mathop{{\mathfrak{so}}}\nolimits_{3}({\mathbb{R}})\cong\mathop{{\mathfrak{so}}}\nolimits_{4}({\mathbb{R}})$ (Lemma 4.3). ∎

Remark 4.9**.**

By Lemma 4.3, the compact dual ${\mathfrak{g}}^{r}$ (Definition 4.1) is not simple if and only if ${\mathfrak{g}}$ is a complex simple Lie algebra.

With Lemma 4.7 and Corollary 4.8, we have largely reduced our problem to the case where ${\mathfrak{g}}$ is simple and not complex, so that the compact dual ${\mathfrak{g}}^{r}$ is simple (Remark 4.9). The final step consists in an inspection of the Riemannian case. We recall from Example 2.6 the reflection $r_{j}(x)=x-2x_{j}e_{j}$ in the hyperplane $e_{j}^{\bot}\subseteq{\mathbb{R}}^{n}$.

Theorem 4.10**.**

Let $({\mathfrak{g}},\tau,\sigma)$ be an irreducible non-abelian dissecting triple such that $({\mathfrak{g}},\tau)$ is Riemannian with $\mathop{{\rm dim}}\nolimits({\mathfrak{g}}/{\mathfrak{h}})=n$. Then $({\mathfrak{g}},\tau,\sigma)$ is equivalent to one of the following types:

• (C)

${\mathfrak{g}}\cong\mathop{{\mathfrak{so}}}\nolimits_{n+1}({\mathbb{R}})$* with $\tau(x)=r_{1}xr_{1}$ and $\sigma(x)=r_{n+1}xr_{n+1}$.*

• (NC)

${\mathfrak{g}}\cong\mathop{{\mathfrak{so}}}\nolimits_{1,n}({\mathbb{R}})$* with $\tau(x)=r_{1}xr_{1}$ and $\sigma(x)=r_{n+1}xr_{n+1}$.*

Proof.

That $({\mathfrak{g}},\tau)$ is Riemannian implies that ${\mathfrak{g}}^{\tau}$ is a compact Lie algebra. If ${\mathfrak{g}}$ is not simple, then Lemma 4.7 implies that ${\mathfrak{g}}\cong\mathop{{\mathfrak{so}}}\nolimits_{3}({\mathbb{R}})\oplus\mathop{{\mathfrak{so}}}\nolimits_{3}({\mathbb{R}})\cong\mathop{{\mathfrak{so}}}\nolimits_{4}({\mathbb{R}})$ with $\tau(x,y)=(y,x)$ and $\sigma(x,y)=(\overline{y},\overline{x})$, which is also equivalent to $\sigma(x,y)=(I_{1,1}yI_{1,1},I_{1,1}xI_{1,1})$ (cf. Remark 4.6). As we have seen in Example 3.4, this implies that (C) holds with $n=3$.

We may therefore assume that ${\mathfrak{g}}$ is simple. That $({\mathfrak{g}},\tau)$ is Riemannian means that either ${\mathfrak{g}}$ is compact or that $\tau$ is a Cartan involution. In the reductive subalgebra ${\mathfrak{h}}^{a}\mathrel{\mathop{\ordinarycolon}}={\mathfrak{g}}^{\sigma\tau}={\mathfrak{h}}_{\mathfrak{l}}\oplus{\mathfrak{q}}_{\mathfrak{m}},$ the one-dimensional subspace ${\mathfrak{q}}_{\mathfrak{m}}$ is central by Lemma 4.5. Let ${\mathfrak{b}}\mathrel{\mathop{\ordinarycolon}}={\mathfrak{z}}_{\mathfrak{g}}({\mathfrak{q}}_{\mathfrak{m}})\cap{\mathfrak{g}}^{-\sigma\tau}$ be the centralizer of ${\mathfrak{q}}_{\mathfrak{m}}$ in ${\mathfrak{q}}^{a}\mathrel{\mathop{\ordinarycolon}}={\mathfrak{g}}^{-\sigma\tau}={\mathfrak{h}}_{\mathfrak{m}}\oplus{\mathfrak{q}}_{\mathfrak{l}}$. Then ${\mathfrak{b}}$ is invariant under ${\mathfrak{h}}^{a}$ and both involutions $\sigma$ and $\tau$. As the Cartan–Killing form $\kappa$ is non-degenerate on $\tau$-invariant subspaces, we have ${\mathfrak{q}}^{a}={\mathfrak{b}}\oplus{\mathfrak{b}}_{1}$, where ${\mathfrak{b}}_{1}$ is the $\kappa$-orthogonal complement of ${\mathfrak{b}}$ in ${\mathfrak{q}}^{a}$. Then the subspace $[{\mathfrak{b}},{\mathfrak{b}}_{1}]\subseteq{\mathfrak{h}}^{a}$ satisfies

[TABLE]

and since $\kappa$ is non-degenerate on the $\tau$-invariant subalgebra ${\mathfrak{h}}^{a}$, we obtain $[{\mathfrak{b}},{\mathfrak{b}}_{1}]=\{0\}$. This implies that ${\mathfrak{b}}+[{\mathfrak{b}},{\mathfrak{b}}]$ is a $\tau$-invariant ideal of ${\mathfrak{g}}$ because it is invariant under ${\mathfrak{h}}^{a}$, ${\mathfrak{b}}$ and ${\mathfrak{b}}_{1}$. Hence the irreducibility of $({\mathfrak{g}},\tau)$ and ${\mathfrak{q}}_{\mathfrak{m}}\not=\{0\}$ entail that ${\mathfrak{b}}=\{0\}$. This shows that

[TABLE]

Hence the one-dimensional subspace ${\mathfrak{a}}\mathrel{\mathop{\ordinarycolon}}={\mathfrak{q}}_{\mathfrak{m}}$ is maximal abelian in ${\mathfrak{q}}$, so that $\mathop{{\rm rank}}\nolimits_{\mathbb{R}}({\mathfrak{g}},\tau)=1$.

Case 1. ${\mathfrak{g}}$ is not compact and $\tau$ is a Cartan involution: Let $0\not=x_{0}\in{\mathfrak{q}}_{\mathfrak{m}}$. Then $\mathop{{\rm ad}}\nolimits x_{0}$ is diagonalizable and, for every eigenvalue $\lambda\in{\mathbb{R}}^{\times}$, the sum

[TABLE]

of the eigenspaces is a subspace of ${\mathfrak{q}}^{a}$, invariant under ${\mathfrak{h}}^{a}$ and $\tau$. The same argument as above now implies that ${\mathfrak{b}}_{\lambda}+[{\mathfrak{b}}_{\lambda},{\mathfrak{b}}_{\lambda}]$ is a $\tau$-invariant ideal of ${\mathfrak{g}}$. As ${\mathfrak{b}}_{\lambda}\not=\{0\}$ by assumption, it follows that ${\mathfrak{q}}^{a}={\mathfrak{b}}_{\lambda}$, so that ${\rm Spec}(\mathop{{\rm ad}}\nolimits x_{0})=\{0,\lambda,-\lambda\}$. We conclude that the restricted root system of $({\mathfrak{g}},\tau)$ is $\Sigma=\{\pm\lambda\}$ of type $A_{1}$. According to the classification of real non-compact simple Lie algebras ([Kn96, pp. 520–543]), this implies that ${\mathfrak{g}}\cong\mathop{{\mathfrak{so}}}\nolimits_{1,n}({\mathbb{R}})$ for $n\geq 2$ and that $M\cong{\mathbb{H}}^{n}$. 333To derive this from the tables in [Hel78], we note that $\mathop{{\mathfrak{su}}}\nolimits^{*}_{4}({\mathbb{C}})\cong\mathop{{\mathfrak{sl}}}\nolimits_{2}({\mathbb{H}})\cong\mathop{{\mathfrak{so}}}\nolimits_{1,5}({\mathbb{R}})$ (cf. [HN12, p. 615]).

To see how $\sigma$ looks like, we recall that it induces a dissecting isometric involution on the hyperbolic space ${\mathbb{H}}^{n}\cong\mathop{{\rm SO}}\nolimits_{1,n}({\mathbb{R}})/\mathop{{\rm SO}}\nolimits_{n}({\mathbb{R}})$ which fixes the base point, say $e_{1}\in{\mathbb{R}}^{1+n}$. As $\mathop{{\rm Iso}}\nolimits({\mathbb{H}}^{n})_{e_{1}}\cong\mathop{\rm O{}}\nolimits_{n}({\mathbb{R}})$ and all dissecting involutions in $\mathop{\rm O{}}\nolimits_{n}({\mathbb{R}})$ are conjugate to $r_{n+1}$, the assertion follows in this case.

Case 2. ${\mathfrak{g}}$ is compact simple: Then ${\mathfrak{g}}^{c}={\mathfrak{h}}\oplus i{\mathfrak{q}}$ is a non-compact simple Lie algebra with Cartan involution $\tau^{c}$, and $({\mathfrak{g}}^{c},\tau^{c},\sigma^{c})$ is dissecting. Therefore Case $1$ implies that ${\mathfrak{g}}^{c}\cong\mathop{{\mathfrak{so}}}\nolimits_{1,n}({\mathbb{R}})$, so that ${\mathfrak{g}}\cong\mathop{{\mathfrak{so}}}\nolimits_{1+n}({\mathbb{R}})$ with $\tau(x)=r_{1}xr_{1}$ and $\sigma(x)=r_{n+1}xr_{n+1}$. ∎

Theorem 4.11**.**

(Classification of dissecting symmetric triples)* Every irreducible dissecting triple $({\mathfrak{g}},\tau,\sigma)$ is equivalent to one of the type $(\mathop{{\mathfrak{so}}}\nolimits_{p,q}({\mathbb{R}}),\tau,\sigma)$, where $\tau$ and $\sigma$ are induced by reflections corresponding to orthogonal anisotropic vectors in ${\mathbb{R}}^{p,q}$.*

Proof.

That the specified triples $(\mathop{{\mathfrak{so}}}\nolimits_{p,q}({\mathbb{R}}),\tau,\sigma)$ are dissecting follows from Example 3.2.

If $({\mathfrak{g}},\tau,\sigma)$ is an irreducible dissecting triple for which ${\mathfrak{g}}$ is not simple, then ${\mathfrak{g}}={\mathfrak{g}}_{1}\oplus{\mathfrak{g}}_{1}$ for ${\mathfrak{g}}_{1}\cong\mathop{{\mathfrak{sl}}}\nolimits_{2}({\mathbb{R}})$ or $\mathop{{\mathfrak{su}}}\nolimits_{2}({\mathbb{C}})$ as in Lemma 4.4. For ${\mathfrak{g}}_{1}\cong\mathop{{\mathfrak{sl}}}\nolimits_{2}({\mathbb{R}})$, we have seen in Example 3.3 that these triples are of the form $(\mathop{{\mathfrak{so}}}\nolimits_{2,2}({\mathbb{R}}),\tau,\sigma)$ as required. If ${\mathfrak{g}}_{1}=\mathop{{\mathfrak{su}}}\nolimits_{2}({\mathbb{C}})$, then $({\mathfrak{g}},\tau)$ is Riemannian, so that Theorem 4.10 implies that $({\mathfrak{g}},\tau,\sigma)$ is equivalent to a triple $(\mathop{{\mathfrak{so}}}\nolimits_{4}({\mathbb{R}}),\tau,\sigma)$ of the required form.

We may therefore assume that ${\mathfrak{g}}$ is simple. If ${\mathfrak{g}}$ is a complex Lie algebra, then Corollary 4.8 implies that ${\mathfrak{g}}^{r}\cong\mathop{{\mathfrak{so}}}\nolimits_{4}({\mathbb{R}})$, and $({\mathfrak{g}}^{r},\tau^{r},\sigma^{r})$ is dissecting by Lemma 4.2. By Theorem 4.10 it is equivalent to a triple with $\tau^{r}(x)=r_{1}xr_{1}$ and $\sigma^{r}(x)=r_{4}xr_{4}$.

If ${\mathfrak{g}}$ is simple and not complex, then $({\mathfrak{g}}^{r},\tau^{r},\sigma^{r})$ is a compact dissecting triple, where ${\mathfrak{g}}^{r}$ is simple (Remark 4.9), hence isomorphic to $(\mathop{{\mathfrak{so}}}\nolimits_{n+1}({\mathbb{R}}),\tau^{r},\sigma^{r})$, where $\tau^{r}(x)=r_{1}xr_{1}$ and $\sigma^{r}(x)=r_{n+1}xr_{n+1}$ (Theorem 4.10). To see how $({\mathfrak{g}},\tau,\sigma)$ looks like, we have to determine all involutions $\theta^{r}$ on ${\mathfrak{g}}^{r}=\mathop{{\mathfrak{so}}}\nolimits_{n+1}({\mathbb{R}})$ commuting with $\tau^{r}$ and $\sigma^{r}$. As any such involution corresponds to an isometric involution on the sphere ${\mathbb{S}}^{n}$, it is induced by an element $\gamma\in\mathop{\rm O{}}\nolimits_{n+1}({\mathbb{R}})$. That $\gamma$ commutes with $\tau^{r}$ means that $\gamma r_{1}\gamma=\pm r_{1}$ (note that $\mathop{{\rm Ad}}\nolimits(-\mathbf{1})=\mathop{{\rm id}}\nolimits_{{\mathfrak{g}}^{r}}$). If $\gamma r_{1}\gamma=-r_{1}$, then $\gamma$ exchanges the two eigenspaces of $r_{1}$. But, as $n>1$, the two eigenspaces of $r_{1}$ have different dimensions, so that we must have $\gamma r_{1}\gamma=r_{1}$, i.e., $\gamma$ commutes with $r_{1}$. We likewise see that $\gamma$ commutes with $r_{n+1}$. Hence $\gamma$ is conjugate under the centralizer of $r_{1}$ and $r_{n+1}$ to an element of the form

[TABLE]

Since $\pm\gamma$ induce the same automorphism on ${\mathfrak{g}}^{r}$, we may assume that $\gamma_{1}=1$. Permuting the entries of $\gamma$, we may now assume that there exists a $p\leq n+1$ such that we have for $q\mathrel{\mathop{\ordinarycolon}}=n-p$

[TABLE]

Depending on $\gamma_{n+1}$, we find the two Lie algebras:

[TABLE]

For $V\mathrel{\mathop{\ordinarycolon}}={\mathbb{R}}^{n+1}$, these Lie algebras can be realized naturally on the real linear subspace $V^{c}\mathrel{\mathop{\ordinarycolon}}=V^{\gamma}\oplus iV^{-\gamma}\subseteq{\mathbb{C}}^{n+1}$, on which they preserve a form of signature $(p+1,q)$, resp., $(p,q+1)$, according to $\mathop{{\rm dim}}\nolimits V^{-\gamma}=q$ and $q+1$, respectively. This completes the proof of the classification theorem. ∎