Pseudo-Riemannian almost quaternionic homogeneous spaces with irreducible isotropy

TL;DR

This paper classifies certain pseudo-Riemannian almost quaternionic homogeneous spaces, showing they are mostly symmetric spaces with some exceptions in lower dimensions, advancing understanding of quaternionic geometry.

Contribution

It proves that such spaces are locally isometric to quaternionic Kähler symmetric spaces in higher dimensions, and provides a counterexample in dimension 12.

Findings

Spaces of dimension ≥16 are symmetric quaternionic Kähler spaces.

In dimension 12, a non-symmetric example exists.

Classification results for pseudo-Riemannian almost quaternionic homogeneous spaces.

Abstract

We show that pseudo-Riemannian almost quaternionic homogeneous spaces with index 4 and an H-irreducible isotropy group are locally isometric to a pseudo-Riemannian quaternionic K\"ahler symmetric space if the dimension is at least 16. In dimension 12 we give a non-symmetric example.

Tables

$A$	$A_0$
$𝔰𝔭(1,n)$	$𝔰𝔭(1,n)$ $\left\{0\right\}$
$𝔰𝔲(1,n)\oplus 𝔲\left(1\right)$	$𝔰𝔲(1,n)\oplus 𝔲\left(1\right)$ $𝔰𝔲(1,n)$ $𝔲\left(1\right)$ $\left\{0\right\}$
$𝔰𝔲(1,n)$	$𝔰𝔲(1,n)$ $\left\{0\right\}$
$𝔰𝔬(1,n)\oplus 𝔰𝔭\left(1\right)$	$𝔰𝔬(1,n)\oplus 𝔰𝔭\left(1\right)$ $𝔰𝔬(1,n)$ $𝔰𝔭\left(1\right)$ $\left\{0\right\}$
$𝔰𝔬(1,n)\oplus 𝔲\left(1\right)$	$𝔰𝔬(1,n)\oplus 𝔲\left(1\right)$ $𝔰𝔬(1,n)$ $𝔲\left(1\right)$ $\left\{0\right\}$
$𝔰𝔬(1,n)$	$𝔰𝔬(1,n)$ $\left\{0\right\}$
$𝔰𝔭\left(1\right)$	$𝔰𝔭\left(1\right)$ $\left\{0\right\}$
$𝔲\left(1\right)$	$𝔲\left(1\right)$ $\left\{0\right\}$
$\left\{0\right\}$	$\left\{0\right\}$

Equations

$$\mathfrak{h}:=\{X+Y\in A\oplus B|\theta (X+A_0)=Y+B_0\}.$$

$$S=\left\{e^{ibt}\left(\begin{array}{cc}\cosh (at)& \sinh (at)\\ \sinh (at)& \cosh (at)\end{array}\right)|t\in \mathbb{R}\right\},$$

$$\mathfrak{sp}(1,n),\mathfrak{u}(1,n),\mathfrak{su}(1,n),\mathfrak{so}(1,n)\oplus \mathfrak{sp}(1)\cdot \text{mathbbm}{1}_{n+1},\mathfrak{so}(1,n)\oplus \mathfrak{u}(1)\cdot \text{mathbbm}{1}_{n+1},$$

$$\mathfrak{so}(1,n),\mathfrak{sp}(1)\cdot \text{mathbbm}{1}_{n+1},\mathfrak{u}(1)\cdot \text{mathbbm}{1}_{n+1},\{0\},$$

$$\mathrm{Sp}(1,n),\mathrm{U}(1,n),\mathrm{SU}(1,n),{\mathrm{SO}}^0(1,n)\left(\mathrm{Sp}(1)\cdot \text{mathbbm}{1}_{n+1}\right),{\mathrm{SO}}^0(1,n)\left(\mathrm{U}(1)\cdot \text{mathbbm}{1}_{n+1}\right),$$

$${\mathrm{SO}}^0(1,n),\mathrm{Sp}(1)\cdot \text{mathbbm}{1}_{n+1},\mathrm{U}(1)\cdot \text{mathbbm}{1}_{n+1}.$$

$$(\mathfrak{so}(1,n)\oplus \mathfrak{sp}(1)\cdot \text{mathbbm}{1}_{n+1},\mathfrak{so}(1,n))\text{and}(\mathfrak{sp}(1)\cdot \text{mathbbm}{1}_{n+1},\{0\}).$$

$$\left[x,y\right]=\underset{\in \mathfrak{h}}{\underbrace{⟨\vec{p},\vec{q}⟩\cdot K(u\times v)-\frac{1}{2}\eta (u,v)\iota (\vec{p}\times \vec{q})}}+\underset{\in {\mathbb{R}}^{1,2}\subset {\mathbb{H}}^{1,2}=\mathfrak{m}}{\underbrace{u\times v(p_0{q}_0-⟨\vec{p},\vec{q}⟩)}},$$

$$\left[x,y\right]$$

$$0$$

Full text

*Keywords: Homogeneous spaces, symmetric spaces, pseudo-Riemannian manifolds, almost quaternionic structures

MSC classification: 53C26, 53C30, 53C35, 53C50

1 Introduction

In [AZ] Ahmed and Zeghib studied pseudo-Riemannian almost complex homogeneous spaces of index $2$ with a $\mathbb{C}$-irreducible isotropy group. They showed that these spaces are already pseudo-Kähler if the dimension is at least $8$. If furthermore the Lie algebra of the isotropy group is $\mathbb{C}$-irreducible then the space is locally isometric to one of five symmetric spaces.

There are two different quaternionic analogues of Kähler manifolds, namely hyper-Kähler and quaternionic Kähler manifolds. In the first case, the complex structure is replaced by three complex structures assembling into a hyper-complex structure $(I,J,K)$, in the second by the more general notion of a quaternionic structure $Q\subset\mathrm{End}\,TM$ on the underlying manifold $M$. Riemannian as well as pseudo-Riemannian quaternionic Kähler manifolds are Einstein and therefore of particular interest in pseudo-Riemannian geometry.

In [CM] the authors investigated the hyper-complex analogue of the topic studied by Ahmed and Zeghib, namely pseudo-Riemannian almost hyper-complex homogeneous spaces of index $4$ with an $\mathbb{H}$-irreducible isotropy group. It turned out that these spaces of dimension greater or equal than $8$ are already locally isometric to the flat space $\mathbb{H}^{1,n}$ except in dimension $12$, where non-symmetric examples exist.

In this article we study the quaternionic analogue, that is we consider pseudo-Riemannian almost quaternionic homogeneous spaces of index $4$ with an $\mathbb{H}$-irreducible isotropy group. The main result of our analysis is the following theorem.

Theorem 1.1**.**

Let $(M,g,Q)$ be a connected almost quaternionic pseudo-Hermitian manifold of index $4$ and $\dim M=4n+4\geq 16$, such that there exists a connected Lie subgroup $G\subset\mathrm{Iso}(M,g,Q)$ acting transitively on $M$. If the isotropy group $H:=G_{p}$, $p\in M$, acts $\mathbb{H}$-irreducibly, then $(M,g,Q)$ is locally isometric to a quaternionic Kähler symmetric space.

Here $\mathrm{Iso}(M,g,Q)$ denotes the subgroup of the isometry group $\mathrm{Iso}(M,g)$ which preserves the almost quaternionic structure $Q$ of $M$. A consequence of the theorem is that the homogeneous space $M$ itself is quaternionic Kähler and locally symmetric. Notice that pseudo-Riemannian quaternionic Kähler symmetric spaces have been classified in [AC]. In Section 3.2 we show, by construction of a non-symmetric example in dimension $12$, that the hypothesis $\dim M\geq 16$ in Theorem 1.1 cannot be omitted. Moreover, we classify in Proposition 3.1 all examples with the same isotropy algebra $\mathfrak{h}=\mathfrak{so}(1,2)\oplus\mathfrak{so}(3)\subset\mathfrak{so}(1,2)\oplus\mathfrak{so}(4)\subset\mathfrak{gl}(\mathbb{R}^{1,2}\otimes\mathbb{R}^{4})\cong\mathfrak{gl}(12,\mathbb{R})$ in terms of the solutions of a system of four quadratic equations for six real variables.

The strategy of the proof of Theorem 1.1 is as follows. We consider the $\mathbb{H}$-irreducible isotropy group $H$ as a subgroup of $\mathrm{Sp}(1,n)\mathrm{Sp}(1)$ and classify the possible Lie algebras. Then we consider the covering $G/H^{0}$ of $M=G/H$ and show by taking into account the possible Lie algebras that it is a reductive homogeneous space. Finally, we show that the universal covering $\tilde{M}$ is a symmetric space. The invariance of the fundamental $4$-form under $G$ then implies that the symmetric space is quaternionic Kähler.

Acknowledgments. This work was partly supported by the German Science Foundation (DFG) under the Collaborative Research Center (SFB) 676 Particles, Strings and the Early Universe.

2 About subgroups of Sp(1,n)Sp(1)

Lemma 2.1** (Goursat’s theorem).**

Let $\mathfrak{g}_{1}$, $\mathfrak{g}_{2}$ be Lie algebras. There is a one-to-one correspondence between Lie subalgebras $\mathfrak{h}\subset\mathfrak{g}_{1}\oplus\mathfrak{g}_{2}$ and quintuples ${\mathcal{Q}}(\mathfrak{h})=(A,A_{0},B,B_{0},\theta)$, with $A\subset\mathfrak{g}_{1}$ $B\subset\mathfrak{g}_{2}$ Lie subalgebras, $A_{0}\subset A$, $B_{0}\subset B$ ideals and $\theta:A/A_{0}\to B/B_{0}$ is a Lie algebra isomorphism.

Proof: Let $\mathfrak{h}\subset\mathfrak{g}_{1}\oplus\mathfrak{g}_{2}$ be a Lie subalgebra and denote by $\pi_{i}:\mathfrak{g}_{1}\oplus\mathfrak{g}_{2}\rightarrow\mathfrak{g}_{i}$, $i=1,2$, the natural projections. Set $A:=\pi_{1}(\mathfrak{h})\subset\mathfrak{g}_{1}$, $B:=\pi_{2}(\mathfrak{h})\subset\mathfrak{g}_{2}$, $A_{0}:=\ker({\pi_{2}}_{|\mathfrak{h}})$ and $B_{0}:=\ker({\pi_{1}}_{|\mathfrak{h}})$. It is not hard to see that $A_{0}$ and $B_{0}$ can be identified with ideals in $A$ and $B$ respectively. Now we can define a map $\tilde{\theta}:A\to B/B_{0}$ as follows. For $X\in A$ take any $Y\in B$ such that $X+Y\in\mathfrak{h}$ and define $\tilde{\theta}(X):=Y+B_{0}$. It is easy to check that this map is well defined. Its kernel is $A_{0}$ so $\tilde{\theta}$ induces a Lie algebra isomorphism $\theta:A/A_{0}\to B/B_{0}$. This defines a map $\mathfrak{h}\mapsto{\mathcal{Q}}(\mathfrak{h})$.

Conversely, a quintuple $Q=(A,A_{0},B,B_{0},\theta)$ as above defines a Lie subalgebra $\mathfrak{h}={\mathcal{G}}(Q)\subset\mathfrak{g}_{1}\oplus\mathfrak{g}_{2}$ by setting

[TABLE]

It is not hard to see that the maps ${\mathcal{G}}$ and ${\mathcal{Q}}$ are inverse to each other. $\hfill\Box$

We will use the following two classification results for $\mathbb{H}$-irreducible subgroups of $\mathrm{Sp}(1,n)$.

Theorem 2.1** ([CM, Corollary 2.1]).**

Let $H\subset\mathrm{Sp}(1,n)$ be a connected and $\mathbb{H}$-irreducible Lie subgroup. Then $H$ is conjugate to one of the following groups:

• $(i)$

$\mathrm{SO}^{0}(1,n)$, $\mathrm{SO}^{0}(1,n)\cdot\mathrm{U}(1)$, $\mathrm{SO}^{0}(1,n)\cdot\mathrm{Sp}(1)$ if $n\geq 2$,

• $(ii)$

$\mathrm{SU}(1,n)$, $\mathrm{U}(1,n)$,

• $(iii)$

$\mathrm{Sp}(1,n)$,

• $(iv)$

$U^{0}=\left\{A\in\mathrm{Sp}(1,1)\ |\ A\mathrm{\Phi}=\mathrm{\Phi}A\right\}\cong\mathrm{Spin}^{0}(1,3)$* with \mathrm{\Phi}=$$\begin{pmatrix}0&-1\\ 1&0\end{pmatrix} if $n=1$.*

Proposition 2.1** ([CM, Proposition 2.4]).**

Let $H\subset\mathrm{Sp}(1,n)$ be an $\mathbb{H}$-irreducible subgroup. Then one of the following is true.

• $(i)$

$H$* is discrete.*

• $(ii)$

$H^{0}=\mathrm{U}(1)\cdot\mathbbm{1}_{n+1}$* or $H^{0}=\mathrm{Sp}(1)\cdot\mathbbm{1}_{n+1}$.*

• $(iii)$

$H^{0}$* is $\mathbb{H}$-irreducible.*

• $(iv)$

$n=1$* and $H^{0}$ is one of the groups $\mathrm{SO}^{0}(1,1)$, $\mathrm{SO}^{0}(1,1)\cdot\mathrm{U}(1)$, $\mathrm{SO}^{0}(1,1)\cdot\mathrm{Sp}(1)$ or*

[TABLE]

for some non-zero real numbers $a,b$.

We denote by $\pi_{1}:\mathfrak{sp}(1,n)\oplus\mathfrak{sp}(1)\to\mathfrak{sp}(1,n)$ and $\pi_{2}:\mathfrak{sp}(1,n)\oplus\mathfrak{sp}(1)\to\mathfrak{sp}(1)$ the canonical projections.

Proposition 2.2**.**

Let $n\geq 2$ and $H\subset\mathrm{Sp}(1,n)\mathrm{Sp}(1)$ be an $\mathbb{H}$-irreducible closed subgroup. Then the Lie algebra $\mathfrak{h}$ is one of the following:

• $(i)$

$\mathfrak{h}=\mathfrak{h}_{0}\oplus\mathfrak{c}$* with $\mathfrak{h}_{0}\in\{\{0\},\mathfrak{so}(1,n)\}$, $\mathfrak{c}\subset\mathfrak{sp}(1)\cdot\mathbbm{1}_{n+1}\oplus\mathfrak{sp}(1)$ and $\pi_{1}(\mathfrak{c})=\mathfrak{sp}(1)\cdot\mathbbm{1}_{n+1}$, $\pi_{2}(\mathfrak{c})=\mathfrak{sp}(1)$, $\mathfrak{c}\cap\mathfrak{sp}(1,n)=\{0\}$, $\mathfrak{c}\cap\mathfrak{sp}(1)=\{0\}$,*

• $(ii)$

$\mathfrak{h}=\mathfrak{h}_{0}\oplus\mathfrak{c}$* with $\mathfrak{h}_{0}\in\{\{0\},\mathfrak{so}(1,n),\mathfrak{su}(1,n)\}$, $\mathfrak{c}\subset\mathfrak{u}(1)\cdot\mathbbm{1}_{n+1}\oplus\mathfrak{u}(1)$ and $\pi_{1}(\mathfrak{c})=\mathfrak{u}(1)\cdot\mathbbm{1}_{n+1}$, $\pi_{2}(\mathfrak{c})=\mathfrak{u}(1)$, $\mathfrak{c}\cap\mathfrak{sp}(1,n)=\{0\}$, $\mathfrak{c}\cap\mathfrak{sp}(1)=\{0\}$,*

• $(iii)$

$\mathfrak{h}=\mathfrak{h}_{0}\oplus\mathfrak{c}$* where $\mathfrak{h}_{0}\subset\mathfrak{sp}(1,n)$ is one of the following Lie algebras*

[TABLE]

[TABLE]

and $\mathfrak{c}\subset\mathfrak{sp}(1)$ is $\{0\},\mathfrak{u}(1)$ or $\mathfrak{sp}(1)$.

Proof: The idea is to apply Goursat’s theorem (Lemma 2.1) to $\mathfrak{h}\subset\mathfrak{sp}(1,n)\oplus\mathfrak{sp}(1)$. The Lie subalgebras $A$, $A_{0}$, $B$ and $B_{0}$ are given by $\pi_{1}(\mathfrak{h})$, $\mathfrak{h}\cap\mathfrak{sp}(1)$, $\pi_{2}(\mathfrak{h})$ and $\mathfrak{h}\cap\mathfrak{sp}(1)$. Let $p:\mathrm{Sp}(1,n)\times\mathrm{Sp}(1)\to\mathrm{Sp}(1,n)$ be the natural projection. Notice that $H\subset\mathrm{Sp}(1,n)\mathrm{Sp}(1)$ is $\mathbb{H}$-irreducible if and only if $p(\hat{H})\subset\mathrm{Sp}(1,n)$ is $\mathbb{H}$-irreducible, where $\hat{H}$ is the preimage of $H$ under the two-fold covering $\mathrm{Sp}(1,n)\times\mathrm{Sp}(1)\to\mathrm{Sp}(1,n)\mathrm{Sp}(1)$. By Proposition 2.1 and Theorem 2.1 we know that $p(\hat{H})$ is either discrete or $(p(\hat{H}))^{0}$ is one of the following subgroups of $\mathrm{Sp}(1,n)$:

[TABLE]

[TABLE]

Since $dp=\pi_{1}$ we immediately obtain all possibilities for $\pi_{1}(\mathfrak{h})$. Furthermore $\mathfrak{h}\cap\mathfrak{sp}(1,n)$ is an ideal of the Lie algebra $\pi_{1}(\mathfrak{h})$. We can read off from the above list a decomposition of $\pi_{1}(\mathfrak{h})$ into ideals, which gives us all possibilities for $\mathfrak{h}\cap\mathfrak{sp}(1,n)$. The resulting list of pairs $(A,A_{0})$ is displayed in a table below.

On the other side there are only three Lie subalgebras of $\mathfrak{sp}(1)$, namely $\mathfrak{sp}(1)$ itself, $\mathfrak{u}(1)$ and $\{0\}$. It follows that $\pi_{2}(\mathfrak{h})$ is one of these three. Again, $\mathfrak{h}\cap\mathfrak{sp}(1)$ is an ideal of $\pi_{2}(\mathfrak{h})$. It follows that the only possibilites for $\mathfrak{h}\cap\mathfrak{sp}(1)$ are the same as for $\pi_{2}(\mathfrak{h})$.

By Goursat’s theorem we have a Lie algebra isomorphism $\theta:A/A_{0}\to B/B_{0}$. Since we know all possibilities for $B$ and $B_{0}$, it follows that $A/A_{0}$ is isomorphic to $\mathfrak{sp}(1)$, $\mathfrak{u}(1)$ or $\{0\}$. Therefore we need to consider all possibilities for $A$ and $A_{0}$, as listed in the following table, and keep only those for which $A/A_{0}$ is isomorphic to $\mathfrak{sp}(1)$, $\mathfrak{u}(1)$ or $\{0\}$.

If $B/B_{0}\cong\mathfrak{sp}(1)$ then $B=\mathfrak{sp}(1)$ and $B_{0}=\{0\}$. The possibilities for $(A,A_{0})$ are

[TABLE]

This gives us case $(i)$. Analogously we get the remaining Lie algebras in $(ii)$ and $(iii)$. $\hfill\Box$

3 Main results

3.1 Proof of the main theorem

Lemma 3.1** ([CM, Lemma 3.1]).**

Let $n\geq 3$ and $\alpha\in\otimes^{3}V^{*}$, where $V=\mathbb{H}^{1,n}$ is considered as real vector space. If $\alpha$ is $\mathrm{SO}^{0}(1,n)$-invariant, then $\alpha=0$.

Remark 3.1**.**

The $\mathrm{SO}^{0}(1,n)$-invariant elements of $\otimes^{3}V^{*}$ are in one-to-one correspondence to the $\mathrm{SO}^{0}(1,n)$-equivariant bilinear maps from $V\times V$ to $V$. It follows from Lemma 3.1 that the corresponding bilinear maps also vanish.

Proof of Theorem 1.1: Let $\rho:H\to\mathrm{GL}(T_{p}M)$ be the isotropy representation. We identify $H$ with its image $\rho(H)$. Since $H$ preserves the metric $g$ and the almost quaternionic structure $Q$, we can consider $H$ as a subgroup of $\mathrm{Sp}(1,n)\mathrm{Sp}(1)$.

In our first step we consider the covering $G/H^{0}$ of $M=G/H$ and show that it is a reductive homogeneous space, i.e. there exists an $H^{0}$-invariant subspace $\mathfrak{m}\subset\mathfrak{g}$ such that $\mathfrak{g}=\mathfrak{h}\oplus\mathfrak{m}$.

We apply Proposition 2.2 to $H^{0}$. The existence of a subspace $\mathfrak{m}$ is clear if $\mathfrak{h}$ is one of the semi-simple Lie algebras in the list. If $\mathfrak{h}$ is one of the abelian Lie algebras contained in $\mathfrak{u}(1)\cdot\mathbbm{1}_{n+1}\oplus\mathfrak{u}(1)$, then the closure of $\mathrm{Ad}(H^{0})\subset\mathrm{GL}(\mathfrak{g})$ is compact and hence there exists an $\mathrm{Ad}(H^{0})$-invariant subspace $\mathfrak{m}$. The remaining Lie algebras in the list have the form $\mathfrak{h}=\mathfrak{s}\oplus\mathfrak{z}$ where $\mathfrak{s}$ is semi-simple containing $\mathfrak{so}(1,n)$ and $\mathfrak{z}$ is the non-trivial centre. Then $\mathfrak{g}$ decomposes into $\mathfrak{g}=\mathfrak{s}\oplus\mathfrak{z}\oplus\mathfrak{m}$ with respect to the action of $\mathfrak{s}$. If we consider the action of $\mathfrak{s}$ on $\mathfrak{m}\cong\mathbb{H}^{1,n}$ as a complex representation, then $\mathfrak{m}$ is either $\mathbb{C}$-irreducible or decomposes into two $\mathbb{C}$-irreducible subrepresentations. Since the elements of $\mathfrak{z}$ commute with $\mathfrak{s}$, they preserve the sum of all non-trivial $\mathfrak{s}$-submodules, which is precisely $\mathfrak{m}$. Thus we have shown that $G/H^{0}$ is a reductive homogeneous space.

Next we show that $\mathfrak{g}=\mathfrak{h}\oplus\mathfrak{m}$ is a symmetric Lie algebra. It is sufficient to show that $\left[\mathfrak{m},\mathfrak{m}\right]\subset\mathfrak{h}$. We restrict the Lie bracket $\left[\cdot,\cdot\right]$ to $\mathfrak{m}\times\mathfrak{m}$ and denote its projection to $\mathfrak{m}$ by $\beta$. It is an antisymmetric bilinear map which is $\mathrm{Ad}(H)$-equivariant. Since $\mathfrak{m}\cong\mathbb{H}^{1,n}$, we can consider $\beta$ as an element of $\otimes^{3}(\mathbb{H}^{1,n})^{*}$. It is also $H^{Zar}$-invariant, where $H^{Zar}$ denotes the Zariski closure. Since $H^{Zar}$ is an algebraic group, it has only finitely many connected components, see [Mi]. Now we show that $(H^{Zar})^{0}$ is non-compact.

Assume that $(H^{Zar})^{0}$ is compact. Since $H^{Zar}$ has only finitely many connected components it follows that $H^{Zar}$ is compact and therefore contained in a maximal compact subgroup of $\mathrm{Sp}(1,n)\mathrm{Sp}(1)$. Hence, $H^{Zar}$ is conjugate to a subgroup of $(\mathrm{Sp}(1)\times\mathrm{Sp}(n))\mathrm{Sp}(1)$ but this contradicts the $\mathbb{H}$-irreducibility of $H^{Zar}$. So we have shown that $(H^{Zar})^{0}$ is non-compact.

Now we apply Proposition 2.2 to $H^{Zar}$. Since $H^{Zar}$ is non-compact we see from the list there that $(H^{Zar})^{0}$ contains $\mathrm{SO}^{0}(1,n)$. Hence, $\beta$ is $\mathrm{SO}^{0}(1,n)$-equivariant. Since $n\geq 3$ it follows from Remark 3.1 that $\beta$ vanishes. This shows that $\mathfrak{g}=\mathfrak{h}\oplus\mathfrak{m}$ is a symmetric Lie algebra and that the universal covering $\tilde{M}=\tilde{G}/\tilde{G}_{p}$ of $M$ is a symmetric space. The fundamental $4$-form $\Omega$ of $\tilde{M}$ is $\tilde{G}$-invariant and since $\tilde{M}$ is a symmetric space $\Omega$ is parallel. In particular $\Omega$ is closed. It is known that for dimension $\geq 12$ an almost quaternionic Hermitian manifold is quaternionic Kähler if $d\Omega=0$, see [S]. This shows that $\tilde{M}$ is furthermore a quaternionic Kähler manifold. Summarizing, we have shown that $M$ is locally isometric to a quaternionic Kähler symmetric space. $\hfill\Box$

3.2 A class of non-symmetric examples in dimension 12

In Theorem 1.1 we did not consider the dimension $12$. This is because the arguments used in the proof to show that $M$ is a reductive homogeneous space do not apply in this dimension, although still $\mathrm{SO}^{0}(1,n)\subset H^{Zar}$ holds. In fact, the proof relies on Lemma 3.1 which holds for dimension $4n+4\geq 16$. If $\dim M=12$ then $n=2$ and then there exist non-trivial anti-symmetric bilinear forms $\mathbb{H}^{1,2}\times\mathbb{H}^{1,2}\to\mathbb{H}^{1,2}$ which are invariant under $\mathrm{SO}^{0}(1,2)$. Therefore in dimension $12$ we cannot be sure if the manifolds are symmetric.

In the following we will give a non-symmetric example by specifying a Lie algebra $\mathfrak{g}=\mathfrak{h}\oplus\mathfrak{m}$ where $\mathfrak{h}$ is a Lie algebra of the list in Proposition 2.2. The pair $(\mathfrak{g},\mathfrak{h})$ defines a simply connected homogeneous space $M=G/H$ where $G$ is a connected and simply connected Lie group with Lie algebra $\mathfrak{g}$ and $H$ is the closed connected Lie subgroup of $G$ with Lie algebra $\mathfrak{h}$.

Let $\mathfrak{h}=\mathfrak{so}(1,2)\oplus\mathfrak{c}$ with $\mathfrak{c}=\{(X\cdot\mathbbm{1}_{3},X)\in\mathfrak{sp}(1)\cdot\mathbbm{1}_{3}\oplus\mathfrak{sp}(1)\ |\ X\in\mathfrak{sp}(1)\}$, see Proposition 2.2 $(i)$. Then we consider the vector space direct sum $\mathfrak{g}:=\mathfrak{h}\oplus\mathfrak{m}$ with $\mathfrak{m}=\mathbb{H}^{1,2}$ and define a Lie bracket on $\mathfrak{g}$ in the following way. For elements $A,B\in\mathfrak{h}$ we take the standard Lie bracket of $\mathfrak{h}$, i.e. $\left[A,B\right]=AB-BA$. Then we define $\left[A,x\right]=-\left[x,A\right]=Ax$ for $A\in\mathfrak{h}$ and $x\in\mathfrak{m}$. Note that, as an $\mathfrak{h}$-module, we can decompose $\mathfrak{m}=\mathbb{H}^{1,2}=\mathbb{R}^{1,2}\otimes\mathbb{H}=\mathbb{R}^{1,2}\otimes\mathbb{R}^{4}$, where the action of $\mathfrak{so}(1,2)$ is by the defining representation on the first factor and trivial on the second and the action of $\mathfrak{c}\cong\mathfrak{so}(3)\subset\mathfrak{so}(4)$ is trivial on the first factor and by the standard four-dimensional representation $\mathbb{H}=\mathbb{R}\oplus\mathrm{Im}\,\mathbb{H}=\mathbb{R}\oplus\mathbb{R}^{3}$ on the second. Finally we have to define the Lie bracket for elements in $\mathfrak{m}=\mathbb{R}^{1,2}\otimes\mathbb{R}^{4}$.

Let $K:\mathbb{R}^{1,2}\to\mathfrak{so}(1,2)$ be an isomorphism of Lie algebras where $\mathbb{R}^{1,2}$ is endowed with the Lorentzian cross product, $\iota:\mathfrak{sp}(1)\to\mathfrak{c}$, $X\to X\cdot\mathbbm{1}_{3}+X$, and let $\eta$ be the standard Lorentz metric on $\mathbb{R}^{1,2}$. Furthermore denote $\left\langle\cdot,\cdot\right\rangle$ the standard inner product on $\mathbb{R}^{4}$. Let $x=u\otimes p$, $y=v\otimes q\in\mathbb{R}^{1,2}\otimes\mathbb{R}^{4}$ and write $p=p_{0}+\vec{p}$, $q=q_{0}+\vec{q}$, where $p_{0},q_{0}\in\mathbb{R}$ and $\vec{p},\vec{q}\in\mathrm{Im}\,\mathbb{H}=\mathbb{R}^{3}$. We set

[TABLE]

where $\vec{p}\times\vec{q}$ is the Euclidian cross product in $\mathrm{Im}\,\mathbb{H}=\mathfrak{sp}(1)$ and $u\times v$ the Lorentzian cross product in $\mathbb{R}^{1,2}$. This extends the partially defined bracket to an anti-symmetric bilinear map $\left[\cdot,\cdot\right]:\mathfrak{g}\times\mathfrak{g}\to\mathfrak{g}$, which satisfies the Jacobi-identity. Hence $\mathfrak{g}$ becomes a Lie algebra. We claim that $(\mathfrak{g},\mathfrak{h})$ is not a symmetric pair. In fact, every $\mathfrak{h}$-invariant complement $\mathfrak{m}^{\prime}$ of $\mathfrak{h}$ in $\mathfrak{g}$ contains $\mathbb{R}^{1,2}\otimes\mathbb{R}^{3}$ (there is no other equivalent $\mathfrak{h}$-submodule in $\mathfrak{g}$) and thus we see from the formula for the bracket that $\left[\mathfrak{m}^{\prime},\mathfrak{m}^{\prime}\right]\nsubseteq\mathfrak{h}$.

For a general classification of the homogeneous spaces with $\mathfrak{h}=\mathfrak{so}(1,2)\oplus\mathfrak{c}$ we need to classify all the Lie algebra structures on the vector $\mathfrak{g}=\mathfrak{h}\oplus\mathbb{R}^{1,2}\otimes\mathbb{R}^{4}$ such that the Lie bracket restricts to the Lie bracket of $\mathfrak{h}$ and to the given representation of $\mathfrak{h}$ on $\mathbb{R}^{1,2}\otimes\mathbb{R}^{4}$. For this one has to describe all the $\mathfrak{h}$-invariant tensors of $\mathrm{\Lambda}^{2}\mathfrak{m}^{*}\otimes\mathfrak{g}\cong\mathrm{\Lambda}^{2}\mathfrak{m}^{*}\otimes\mathfrak{h}\oplus\mathrm{\Lambda}^{2}\mathfrak{m}^{*}\otimes\mathfrak{m}$ which satisfy the Jacobi-identity. With the above notation, these bilinear maps have the following form

[TABLE]

where $a,b,c,d,a_{1},a_{2},a_{3}\in\mathbb{R}$. The bracket satisfies the Jacobi-identity if and only if the following equations hold

[TABLE]

Summarizing we obtain the following proposition.

Proposition 3.1**.**

Every solution $(a,b,c,a_{1},a_{2},a_{3})$ of the quadratic system (1)-(4) defines a connected and simply connected homogeneous almost quaternionic pseudo-Hermitian manifold $G/H$ with isotropy algebra $\mathfrak{h}=\mathfrak{so}(1,2)\oplus\mathfrak{so}(3)\subset\mathfrak{so}(1,2)\oplus\mathfrak{so}(4)\subset\mathfrak{gl}(\mathbb{R}^{1,2}\otimes\mathbb{R}^{4})\cong\mathfrak{gl}(12,\mathbb{R})$. Conversely, every such homogeneous space arises by this construction.

The above example corresponds to $a=0$, $b=1$, $c=-\frac{1}{2}$, $d=0$, $a_{1}=1$, $a_{2}=-1$ and $a_{3}=0$.