Conformal properties of indefinite bi-invariant metrics

TL;DR

This paper investigates when indefinite bi-invariant metrics on Lie groups are conformally equivalent to Einstein metrics, providing a complete classification in certain Lorentzian and (2,n-2) signatures.

Contribution

It offers a comprehensive analysis of conformally Einstein properties of metric Lie algebras across various cases, extending known results to indefinite metrics.

Findings

Simple Lie algebras are conformally Einstein iff they are Einstein or isomorphic to sl(2,C) and conformally flat.

Double extensions by higher rank simple Lie algebras are not conformally Einstein.

Oscillator algebras are conformally Einstein, but their double extensions are not.

Abstract

An indecomposable Lie group with Riemannian bi-invariant metric is always simple and hence Einstein. For indefinite metrics this is no longer true, not even for simple Lie groups. We study the question of whether a semi-Riemannian bi-invariant metric is conformal to an Einstein metric. We obtain results for all three cases in the structure theorem by Medina and Revoy for indecomposable metric Lie algebras: the case of simple Lie algebras, and the cases of double extensions of metric Lie algebras by $\mathbb{R}$ or a simple Lie algebra. Simple Lie algebras are conformally Einstein precisely when they are Einstein, or when equal to $\mathfrak{sl}_2\mathbb{C}$ and conformally flat. Double extensions of metric Lie algebras by simple Lie algebras of rank greater than one are never conformally Einstein, and neither are double extensions of Lorentzian oscillator algebras, whereas the oscillator algebras themselves are conformally Einstein. Our results give a complete answer to the question of which metric Lie algebras in Lorentzian signature and in signature (2,n-2) are conformally Einstein.

Equations

$$\mathsf{P}-\nabla {\rm Y}+{{\rm Y}}^2$$

$$\mathfrak{n}=\{X\in \mathfrak{g}\mid X\text{makebox}[7.0pt]\mathrm{C}=0\}\subset \mathfrak{g},$$

$$\mathrm{R}(X,Y)Z$$

$$\mathrm{Ric}(X,Y)$$

$${\mathrm{R}}_{ijkl}=\mathrm{R}(X_i,X_j,X_k,X_l),{\mathrm{R}}_{ij}=\mathrm{Ric}(X_i,X_j)={\mathrm{R}}_{kij}^k={\mathrm{g}}^{kl}{\mathrm{R}}_{kijl}.$$

$$(n-2)\mathsf{P}=\mathrm{Ric}-\frac{\rho }{2(n-1)}\mathrm{g},$$

$${\mathrm{C}}_{abcd}={\mathrm{R}}_{abcd}+2({\mathrm{g}}_{a[c}{\mathsf{P}}_{d]b}+{\mathrm{g}}_{b[d}{\mathsf{P}}_{c]a}),$$

$$\mathsf{J}={\mathrm{g}}^{bc}{\mathsf{P}}_{bc}=\frac{\rho }{2(n-1)}.$$

$${\mathrm{A}}_{abc}={\nabla }_{[b}{\mathsf{P}}_{c]a},{\mathrm{B}}_{bc}={\mathsf{P}}^{ad}{\mathrm{C}}_{abcd}-{\nabla }^d{\mathrm{A}}_{bcd}.$$

$$\hat{\mathsf{P}}=\mathsf{P}-\nabla {\rm Y}+{{\rm Y}}^2-\frac{1}{2}\mathrm{g}({\rm Y},{\rm Y})\mathrm{g},\hat{\mathsf{J}}={\mathrm{e}}^{-2\phi }\left(\mathsf{J}-\mathrm{div}({\rm Y})-\frac{n-2}{2}\mathrm{g}({\rm Y},{\rm Y})\right),$$

$$\mathsf{P}-{\nabla }^2\phi +(\mathrm{d}\phi {)}^2-\frac{\mathsf{J}-\Delta (\phi )+\mathrm{d}\phi (\nabla \phi )}{n}\mathrm{g}=0,$$

$$\mathsf{P}-\nabla {\rm Y}+{{\rm Y}}^2=\lambda \mathrm{g},$$

$$\lambda =\frac{1}{n}\left(\mathsf{J}-\mathrm{div}({\rm Y})+\mathrm{g}({\rm Y},{\rm Y})\right).$$

$${\nabla }^2\sigma +\sigma \mathsf{P}=\mu \mathrm{g},$$

$$\mathrm{A}-\mathrm{C}(V,.,.,.)=0,$$

$$\mathrm{B}-(n-4)\mathrm{C}(V,.,.,V)=0.$$

$${\nabla }_{X}Y=\frac{1}{2}[X,Y],\mathrm{R}(X,Y)Z=-\frac{1}{4}\left[\left[X,Y\right],Z\right],\mathrm{Ric}=-\frac{1}{4}K,$$

$${\mathrm{ad}}_X\mathrm{R}=0,{\mathrm{ad}}_X\mathrm{Ric}=0,\text{ etc.,}$$

$$\mathfrak{n}=\mathfrak{ann}(\mathrm{C})=\{X\in \mathfrak{g}\mid X\text{makebox}[7.0pt]\mathrm{C}=0\}\subset \mathfrak{g}$$

$$0={\mathrm{R}}_i^k{\mathrm{R}}_{kj}={\mathrm{R}}_i^k{\mathrm{R}}_j^l{\mathrm{g}}_{kl},$$

$${\mathrm{Ric}}^2(X,Y)=-\frac{1}{16}{\mathrm{tr}}^{\mathrm{g}}(K({\mathrm{ad}}_X(.),{\mathrm{ad}}_Y(.))={\mathrm{tr}}_{(1,3),(2,6)}^{\mathrm{g}}(\mathrm{Ric}\otimes \mathrm{R})(X,Y),$$

$${\mathrm{R}}_a^p{\mathrm{R}}_{pb}=-\frac{1}{16}{\mathrm{g}}^{kl}{c}_{ak}^p{c}_{bl}^q{K}_{pq}={\mathrm{R}}^{pq}{\mathrm{R}}_{pabq},$$

$${\mathrm{Ric}}^2(X_i,X_j)$$

$${\mathrm{R}}^{kl}{\mathrm{R}}_{kijl}$$

$$(n-2{)}^2\mathrm{B}=n{\mathrm{Ric}}^2-\frac{n\rho }{n-1}\mathrm{Ric}+\left(\frac{{\rho }^2}{n-1}-\mathrm{tr}({\mathrm{Ric}}^2)\right)\mathrm{g},$$

$$(n-2{)}^2{\mathrm{B}}_{bc}=(n-2){\mathrm{R}}^{ad}{\mathrm{R}}_{abcd}+2{\mathrm{R}}_{bd}{\mathrm{R}}_c^d-\frac{n\rho }{n-1}{\mathrm{R}}_{bc}+\left(\frac{{\rho }^2}{n-1}-{\mathrm{R}}_{ad}{\mathrm{R}}^{ad}\right){\mathrm{g}}_{bc}.$$

$$0=A^2+bA+c1,\text{ for b,c∈R.}$$

$$b=-2\lambda ,c={\lambda }^2.$$

$$b=-\lambda -\mu ,c=\lambda \mu .$$

$$0={\mu }^2+b\mu +c=(\lambda -\mu {)}^2,$$

$$0={\mathrm{Ric}}^2+b\mathrm{Ric}+c\mathrm{g},\text{ with b=−(n−1)ρ and c=n1(n−1ρ2−tr(Ric2)).}$$

Full text

The second author acknowledges support from the Australian Research Council via the grant FT110100429. The first author was funded by a Master of Philosophy scholarship from the University of Adelaide.

1. 1 Introduction and statement of results
2. 2 Conformal Einstein metrics
3. 3 Bi-invariant metrics on Lie groups
4. 4 Simple metric Lie algebras
5. 5 Metric Lie algebras via double extensions
6. 6 The oscillator algebras and related double extensions

1. Introduction and statement of results

In this paper, we study the conformal properties of bi-invariant metrics on Lie groups by considering properties of their Lie algebras. Recall that a semi-Riemannian metric on a Lie group $G$ is called bi-invariant if all multiplications from left and right are isometries. A Lie group with a bi-invariant semi-Riemannian metric is called a metric Lie group. A bi-invariant metric of signature $(p,q)$ on $G$ induces a scalar product of signature $(p,q)$ on the Lie algebra $\mathfrak{g}$ of $G$ that is $\mathrm{Ad}$-invariant, i.e., invariant under the adjoint representation $\mathrm{Ad}$ of $G$ on $\mathfrak{g}$, and consequently invariant under its differential $\mathrm{ad}$. In fact, on a connected Lie group, bi-invariant metrics are in 1-1 correspondence with $\mathrm{ad}$-invariant scalar products on the Lie algebra. A Lie algebra $\mathfrak{g}$ with a scalar product (of signature $(p,q)$) is called a metric Lie algebra (of signature $(p,q)$).

In the following, when studying the geometry of a metric Lie group $G$ we will do this by studying metric Lie algebras, and when referring to geometric objects on $G$, such as the curvature tensor, the Ricci tensor, etc., we will just refer to curvature tensor, Ricci tensor, etc. of the metric Lie algebra $\mathfrak{g}$.

A metric Lie algebra is called decomposable if the Lie algebra is isomorphic to a direct sum of orthogonal ideals. If there is no such decomposition, we call the metric Lie algebra and its corresponding Lie group indecomposable. Due to splitting theorems of de Rham and Wu decomposability in this algebraic sense is related to decomposability in the geometric sense, i.e., to the fact that the metric Lie group, if it is simply connected, decomposes into a semi-Riemannian product manifold. Hence, indecomposable metric Lie algebras can be considered as the fundamental building blocks of metric Lie algebras. More surprising is the following striking structure result for indecomposable metric Lie algebras:

Theorem 1.1** (Medina & Revoy [22]).**

Every indecomposable metric Lie algebra is either one-dimensional, simple, or a double extension of a metric Lie algebra $\mathfrak{h}\not=\{0\}$ by another Lie algebra $\mathfrak{s}$ and a Lie algebra homomorphism $\delta:\mathfrak{s}\to\mathfrak{so}(\mathfrak{h})\cap{\mathfrak{der}}(\mathfrak{h})$ into to the skew derivations of $\mathfrak{h}$ such that:

1. (a)

$\mathfrak{s}$* is simple or $\mathfrak{s}=\mathbb{R}$, and* 2. (b)

the image of $\delta$ is not contained in the inner derivations $\mathrm{ad}(\mathfrak{h})$, **[10, 11]**.

For a double extension, the signature of the metric is $(p+\dim\mathfrak{s},q+\dim\mathfrak{s})$, where $(p,q)$ is the signature of the metric Lie algebra $\mathfrak{h}$.

It is known that indecomposable Riemannian metric Lie algebras are simple or one-dimensional; the signature description from Theorem 1.1 confirms this. The interesting part of Theorem 1.1 however is the statement about indefinite metric Lie algebras and these will be a focus of our paper.

In the following we will study conformal properties of metric Lie groups, such as (local) conformal flatness or being (locally) conformally equivalent to an Einstein space (for precise definitions see the following paragraph and Section 2). As we will formulate our results in terms of the associated metric Lie algebra, we leave aside the difficulties arising from the transition from local to global and the fact that there may be several (locally isometric) metric Lie groups having the same metric Lie algebra. We will say that a metric Lie algebra $\mathfrak{g}$ has a conformal property if a Lie group with bi-invariant metric and metric Lie algebra $\mathfrak{g}$ has the corresponding local conformal property. For example, given the equivalence of local conformal flatness with the vanishing of the Weyl tensor, we say that a metric Lie algebra $\mathfrak{g}$ is conformally flat if its Weyl tensor vanishes.

Our main focus is the property of a bi-invariant metric on a Lie group to be locally conformally equivalent to an Einstein metric (again see Section 2 for details). Of course, the resulting metric is no longer bi-invariant (unless the scaling function is constant). In contrast to the locally conformally flat property, there is in general no tensorial condition that is equivalent to the locally conformal Einstein property. Instead, it is equivalent to the following differential equation: a metric $\mathrm{g}$ on a manifold $M$ is locally conformally equivalent to an Einstein metric if and only if

1. (A)

each point in $M$ has a neighbourhood $U$ with a closed $1$-form $\Upsilon\in\Gamma(T^{*}U)$ such that that the trace-free part of

[TABLE]

vanishes, where $\mathsf{P}$ is the Schouten tensor and $\nabla$ the Levi-Civita connection of $\mathrm{g}$.

For Lie groups with bi-invariant metric this property cannot be be formulated purely in terms of the metric Lie algebra. Instead we make the following definition:

Definition 1.1**.**

A metric Lie algebra $\mathfrak{g}$ is conformally Einstein if for the unique simply connected metric Lie group $(G,\mathrm{g})$ with metric Lie algebra $\mathfrak{g}$ the bi-invariant metric $\mathrm{g}$ is locally conformally equivalent to an Einstein metric111Of course, as we only require the local property in the definition, if the simply connected metric Lie group is locally conformally Einstein, then all other metric Lie groups with the same metric Lie algebra also satisfy this property..

Even though in general the locally conformally Einstein property is not fully characterised by a tensorial condition, there are certain tensorial obstructions for the original metric to be conformally Einstein. Remarkably, under some genericity conditions on the Weyl tensor, the vanishing of these obstructions is not only necessary but also sufficient for the metric to be conformally Einstein. In general dimensions these obstructions were found by Gover and Nurowski [14]. They allow us to check effectively whether a given metric can be conformally Einstein by computing certain tensors, instead of attempting to solve the PDE in (A) directly. We will see that in the case of a metric Lie algebra $\mathfrak{g}$, the vanishing of these obstructions simplifies to the following two conditions:

1. (B)

The Bach tensor $\mathrm{B}$ of $\mathfrak{g}$ vanishes, and 2. (C)

if $n=\dim(\mathfrak{g})>4$ and $\mathfrak{g}$ is not already Einstein, the Weyl nullity ideal $\mathfrak{n}$ in $\mathfrak{g}$ is not zero. Here $\mathfrak{n}$ is the ideal

[TABLE]

where $\mathrm{C}$ is the Weyl tensor of $\mathfrak{g}$.

Note that the question whether an indecomposable Riemannian, and hence simple, metric Lie algebra is conformally Einstein is trivial as it is already Einstein: the only candidate for an $\mathrm{ad}$-invariant positive definite bilinear form is the Killing form, and therefore the simple Lie algebra $\mathfrak{g}$ is of compact type and Einstein with positive Einstein constant (see for example Milnor’s classical paper [23]).

With the aim of determining which indefinite metric Lie algebras satisfy the necessary conditions (B) and (C), we start by describing the Bach tensor of a metric Lie algebra and expressing its vanishing in terms of the Ricci tensor. In Section 3 we prove our first result:

Theorem 1.2**.**

A metric Lie algebra of dimension $n>2$ is Bach flat if and only if it is Einstein, or its Ricci tensor $\mathrm{Ric}$ satisfies one of the following conditions:

1. (1)

$\mathrm{Ric}$* is $2$-step nilpotent. * 2. (2)

$\mathrm{Ric}$* is diagonalisable with two different eigenvalues $\lambda\not=0$ and $\mu=-\tfrac{n-k-1}{k-1}\lambda$, where $k$ is the dimension of the eigenspace of $\lambda$, and with non-degenerate eigenspaces. In this case, if $\mathrm{Ric}$ has a non trivial kernel, it is of dimension $1$ and non degenerate.*

By a $2$-step nilpotent Ricci tensor we mean that the endomorphism of $\mathfrak{g}$ that is obtained by dualising the Ricci tensor with the metric squares to zero, $\mathrm{Ric}^{2}=0$ (including the Ricci-flat case). As a corollary we obtain that solvable metric Lie algebras are Bach flat.

Next, in Section 4 we consider the case of simple metric Lie algebras $\mathfrak{g}$ in Theorem 1.1 by considering their complexifications $\mathfrak{g}^{\mathbb{C}}$. Extending the trivial Riemannian situation, we show:

Theorem 1.3**.**

Let $\mathfrak{g}$ be a simple metric Lie algebra. If $\mathfrak{g}^{\mathbb{C}}$ is simple, the metric is given by the Killing form and hence Einstein. If $\mathfrak{g}^{\mathbb{C}}$ is not simple, then there is a $2$-parameter family of bi-invariant metrics of neutral signature $(\tfrac{n}{2},\tfrac{n}{2})$. Moreover:

1. (1)

The only Bach flat metrics in this family are the multiples of the Killing form of $\mathrm{g}$ and of the imaginary part of the Killing form of $\mathrm{g}^{\mathbb{C}}$. 2. (2)

The only conformal Einstein metrics in this class are multiples of the Killing form of $\mathrm{g}$ (which are Einstein) and, when $\mathfrak{g}=\mathfrak{sl}_{2}\mathbb{C}$ (as real Lie algebra), the multiples of the imaginary part of the Killing form of $\mathfrak{sl}_{2}\mathbb{C}$ (which are conformally flat).

Motivated by the other cases in Theorem 1.1 we turn to metric Lie algebras that are given by double extensions (for definition and details see Section 5.1). We obtain the following consequence of Theorem 1.2.

Corollary 1.1**.**

Let $\mathfrak{g}$ be a metric Lie algebra given by a double extension as in Definition 5.1. Then

1. (1)

$\mathfrak{g}$* is Einstein if and only if it is Ricci flat, and* 2. (2)

$\mathfrak{g}$* is Bach-flat if and only if $\mathrm{Ric}^{2}=0$.*

As another consequence to Theorem 1.2 we obtain that double extensions $\mathfrak{g}$ of nilpotent Lie algebras $\mathfrak{h}$ have $2$-step nilpotent Ricci tensor.

Next we turn to the case in Theorem 1.1 where $\mathfrak{g}$ is the double extension of a metric Lie algebra by a simple Lie algebra. Using the the necessary conditions (B) and (C) we show in Section 5.2:

Theorem 1.4**.**

If the double extension $\mathfrak{g}$ of a metric Lie algebra $\mathfrak{h}$ by a simple Lie algebra $\mathfrak{s}$ is conformally Einstein, then $\mathfrak{s}=\mathfrak{sl}_{2}\mathbb{R}$ or $\mathfrak{s}=\mathfrak{so}(3)$.

While we will leave the cases of $\mathfrak{s}=\mathfrak{sl}_{2}\mathbb{R}$ or $\mathfrak{s}=\mathfrak{so}(3)$ undecided in general, in Example 5.1 we define a double extension of $\mathfrak{h}=\mathbb{R}^{3}$ by $\mathfrak{s}=\mathfrak{so}(3)$ that satisfies both conditions (B) and (C), but does not satisfy condition (A) and hence is not conformally Einstein.

The remaining case in Theorem 1.1 of double extensions $\mathfrak{g}$ by the $1$-dimensional Lie algebra $\mathfrak{s}=\mathbb{R}$ is the most difficult one. It is however important as every indecomposable Lorentzian metric Lie algebra is either isomorphic to $\mathfrak{sl}_{2}\mathbb{R}$, or to a Lorentzian oscillator algebra, [21]. In general, the oscillator algebra $\mathfrak{osc}_{\Phi}(t,s)$ of signature $(t+1,s+1)$ is defined as the double extension of the abelian metric Lie algebra $\mathbb{R}^{t,s}$ by $\mathbb{R}$ and $\Phi\in\mathfrak{so}(t,s)$. The Lorentzian oscillator algebras of dimension $n$ then are denoted by $\mathfrak{osc}_{\Phi}(n-2)=\mathfrak{osc}_{\Phi}(0,n-2)$. All oscillator algebras are solvable and hence Bach flat. In Section 6.1 we show:

Theorem 1.5**.**

The oscillator Lie algebras $\mathfrak{osc}_{\Phi}(t,s)$ are conformally Einstein. In particular, all indecomposable Lorentzian metric Lie algebras and all indecomposable $4$-dimensional metric Lie algebras are conformally Einstein.

The statement about the Lorentzian metric Lie algebras in this theorem is already known from a more general result in [20], where it was shown that Lorentzian plane waves and hence Lorentzian symmetric spaces with solvable transvection group, the so-called Cahen-Wallach spaces [6], are conformally Einstein. In fact, locally they admit two different rescalings to Ricci-flat metrics.

Finally we turn to indecomposable metric Lie algebras $\mathfrak{g}$ of signature $(2,n-2)$, of which there are three nonsimple types [4]: the oscillator algebras $\mathfrak{g}=\mathfrak{osc}_{\Phi}(1,n-3)$, which are conformally Einstein by Theorem 1.5, and the double extensions $\mathfrak{g}$ of $\mathfrak{h}$ by $\mathbb{R}$ and a derivation of $\mathfrak{h}$, for either $\mathfrak{h}=\mathfrak{osc}_{\Phi}(n-4)$ or $\mathfrak{h}=\mathfrak{osc}_{\Phi}(n-5)\oplus\mathbb{R}$. It turns out that both of the latter are Bach flat but are not conformally Einstein. This and checking that no simple real Lie algebra has a Killing form of signature $(2,n-2)$ for $n\geq 4$ (for example in [26, Chapter 15]), yields our final result.

Theorem 1.6**.**

The oscillator algebras $\mathfrak{osc}_{\Phi}(1,n-3)$, with $\Phi\in\mathfrak{so}(1,n-3)$, are the only indecomposable metric Lie algebras of signature $(2,n-2)$ that are conformally Einstein.

The results in signature $(2,n-2)$ indicate that there might not be a general pattern for the conformally Einstein property in higher signature, however the methods and results presented here should allow one to decide in specific cases whether or not the property holds.

Acknowledgements

Many results in this paper were obtained as part the first author’s Master thesis [12], which was written at the University of Adelaide under supervision of Michael Murray and the second author. We would like to thank Michael Murray for his support and helpful remarks. We also thank Vicente Cortés and Wolfgang Globke for useful discussions on some aspects of the paper, and Michael Eastwood for crucial comments on Example 5.1.

2. Conformal Einstein metrics

Let $(M,\mathrm{g})$ be a semi-Riemanian manifold. Our conventions regarding the Riemann curvature tensor $\mathrm{R}$ and the Ricci curvature $\mathrm{Ric}$ are as follows:

[TABLE]

We will frequently use index notation as abstract indices (in the sense of Penrose) as well as denoting the component of tensor with respect to a basis $X_{i}$, e.g.,

[TABLE]

We will also dualise tensors with the metric, i.e., raising and lowering indices using the metric and its inverse. Moreover, we define the Schouten tensor $\mathsf{P}$ by

[TABLE]

where $\rho$ is the scalar curvature, and the Weyl tensor as

[TABLE]

where $T_{[ab]}=\frac{1}{2}(T_{ab}-T_{ba})$ denotes the skew-symmetrisation of a tensor. The trace of the Schouten tensor and the scalar curvature $\rho=\mathrm{g}^{bc}\mathrm{R}_{bc}$ are related by

[TABLE]

Finally, the Cotton and Bach tensors are given by

[TABLE]

Finally, a semi-Riemannian manifold $(M,\mathrm{g})$ is Einstein and $\mathrm{g}$ is an Einstein metric if its Ricci tensor, or equivalently its Schouten tensor, is a (possibly vanishing) multiple of the metric. If $\mathrm{Ric}=0$, the metric is Ricci-flat.

Given a smooth manifold $M$ we say that two semi-Riemannian metrics $\mathrm{g}$ and $\widehat{\mathrm{g}}$ on $M$ are conformally equivalent if there is a smooth function $\varphi$ on $M$ such that $\widehat{\mathrm{g}}=\mathrm{e}^{2\varphi}\mathrm{g}$, and that they are locally conformally equivalent if each point in $M$ has a neighbourhood $U$ such that on $U$ the metrics $\widehat{\mathrm{g}}|_{U}$ and $\mathrm{g}|_{U}$ are conformally equivalent. The Schouten tensors $\mathsf{P}$ and $\widehat{\mathsf{P}}$ of $\mathrm{g}$ and $\widehat{\mathrm{g}}$ and their traces $\mathsf{J}$ and $\widehat{\mathsf{J}}$ are then related as follows,

[TABLE]

where $\Upsilon={\mathrm{d}}\varphi$ (see for example [5]). Moreover, a semi-Riemannian manifold is (locally) conformally flat if it (locally) conformally equivalent to the flat metric. It is well known since [9, p. §28] that local conformal flatness is equivalent to the vanishing of the Weyl tensor $\mathrm{C}$. Note that, even for simply connected manifolds, $\mathrm{C}=0$ does not imply global conformal flatness, as the example of the sphere shows.

A semi-Riemannian manifold $(M,\mathrm{g})$ is (locally) conformally Einstein if $\mathrm{g}$ is (locally) conformally equivalent to an Einstein metric. The transformation of the Schouten tensor under $\mathrm{g}\mapsto\widehat{\mathrm{g}}$ in eq. 2.1 reveals that the metric $\widehat{\mathrm{g}}=\mathrm{e}^{2\varphi}\mathrm{g}$ is Einstein if and only if the function $\varphi$ on $M$ satisfies the following PDE

[TABLE]

where $\nabla\varphi$ is the gradient, $\nabla^{2}\varphi=\nabla{\mathrm{d}}\varphi$ denotes the Hessian, $\Delta(\varphi)=\mathrm{div}(\nabla\varphi)$ the Laplacian of $\varphi$ and $\mathsf{J}$ is the trace of $\mathsf{P}$, all with respect to $\mathrm{g}$. Therefore, locally the following conditions are equivalent:

1. (1)

$(M,\mathrm{g})$ is locally conformally Einstein; 2. (2)

each point in $M$ has a neighbourhood $U$ with a function $\varphi\in C^{\infty}(U)$ that solves the PDE (2.2); 3. (3)

each point in $M$ has a neighbourhood $U$ with a closed $1$-form $\Upsilon\in\Gamma(T^{*}U)$ that satisfies

[TABLE]

for some function $\lambda$.

In fact, if $\Upsilon$ solves eq. 2.3 for some $\lambda$, then $\lambda$ is determined by

[TABLE]

Remark 2.1**.**

When substituting $\varphi=-\log\sigma$, the PDE (2.2) is equivalent to the linear PDE for $\sigma$,

[TABLE]

for some function $\mu$ determined by taking the trace of the equation. This enables the prolongation of this equation that leads to another equivalence to the local conformal Einstein property: the metric $\widehat{\mathrm{g}}=\sigma^{-2}\mathrm{g}$ is an Einstein metric if and only if $\sigma$ has no zeros and satisfies eq. 2.4, which in turn is equivalent to $\left(\sigma,{\mathrm{d}}\sigma,-\tfrac{1}{n}(\Delta\sigma+\sigma\mathsf{P})\right)$ being a parallel section of the normal conformal tractor bundle, see [3, 7]. Solutions to eq. 2.4 may however have zeros along sets of measure zero, so their existence is not equivalent to the local conformally Einstein property on all of $M$ but only on a dense open subset, see for example [13].

Analysing the transformation of the Cotton tensor and the Bach tensor, Gover and Nurowski derived the following necessary conditions for a metric to be conformally Einstein.

Theorem 2.1** (Gover & Nurowski [14, Proposition 2.1]).**

Let $(M,\mathrm{g})$ be a semi-Riemannian manifold of dimension $n\geq 3$ that is locally conformally Einstein with Cotton, Bach and Weyl tensors $\mathrm{A}$, $\mathrm{B}$ and $\mathrm{C}$ respectively. Then there is a $V\in TM$ such that

[TABLE]

and

[TABLE]

If $(M,\mathrm{g})$ is not already Einstein, then $V$ does not vanish identically.

In this theorem, $V$ corresponds to the metric dual of the closed $1$-form $\Upsilon$ in eq. 2.3, and hence can locally be realised as a gradient vector field. Gover and Nurowski have in fact shown a much stronger result [14, Theorem 2.2]: under a certain genericity condition on the Weyl tensor, the eqs. 2.5 and 2.6 imply that $V$ locally is a gradient vector field and hence that $\mathrm{g}$ is locally conformally Einstein. In this article we will not use this stronger version, because the genericity condition is not satisfied for the class of bi-invariant metrics we will consider. Instead, we mainly evaluate the conditions (2.5) and (2.6) on metric Lie algebras.

3. Bi-invariant metrics on Lie groups

Let $G$ denote a metric Lie group and $\mathfrak{g}$ the corresponding metric Lie algebra with $\mathrm{ad}$-invariant scalar product $\langle.,.\rangle$, which we will also denote by $\mathrm{g}$ when emphasising that it induces a metric or when using index notation. We denote the Levi-Civita connection of $\mathrm{g}$ by $\nabla$. For elements $X,Y,Z\in\mathfrak{g}$, we have

[TABLE]

where $K$ is the Killing form of $\mathfrak{g}$, and $[X,Y]$ is the Lie bracket on $\mathfrak{g}$. All of the curvature tensors are $\mathrm{ad}$-invariant, that is

[TABLE]

and hence are parallel with respect to $\nabla$. In particular, the scalar curvature is constant and the Cotton tensor vanishes, $\mathrm{A}=0$. Moreover, Lie algebra elements that annihilate one of these tensors form an ideal. For example, for the Weyl tensor $\mathrm{C}$,

[TABLE]

forms an ideal. We call this ideal the Weyl nullity ideal.

From the conditions in Theorem 2.1 we get the following necessary conditions for a metric Lie algebra to be conformally Einstein:

Corollary 3.1**.**

If a metric Lie algebra $\mathfrak{g}$ is conformally Einstein, then it is Bach flat and, if $\mathfrak{g}$ is not Einstein, the Weyl nullity ideal is not zero, $\mathfrak{n}\not=\{0\}$.

Proof.

If $\mathfrak{g}$ is conformally Einstein in the sense of Definition 1.1, the corresponding simply connected Lie group $G$ is locally conformally Einstein. This implies that locally there is a closed one form $\Upsilon$ that satisfies eq. 2.3. Hence by Theorem 2.1 the obstructions in (2.5) and (2.6) vanish for a tangent vector $V\in TG$. Since $\nabla\mathsf{P}=0$ the Cotton tensor vanishes and so condition (2.5) reduces to $V\makebox[7.0pt]{\rule{6.0pt}{0.3pt}\rule{0.3pt}{5.0pt}}\,\mathrm{C}=0$. Hence, condition (2.6) implies that $\mathrm{B}=0$. Moreover, if $\mathfrak{g}$ is not already Einstein, there is a point in $p\in G$ such that $V|_{p}\not=0$. Let $0\not=X\in\mathfrak{g}$ such that $X|_{p}=V|_{p}$. Then the invariance of $\mathrm{C}$ shows that $X\makebox[7.0pt]{\rule{6.0pt}{0.3pt}\rule{0.3pt}{5.0pt}}\,\mathrm{C}=0$, i.e., that there is a nonzero element in the Weyl nullity ideal. ∎

Before studying the Weyl nullity ideal for metric Lie algebras in the next sections, we analyse Bach flatness. A crucial tensor for this is the square of the Ricci endomorphism, that is, the composition of the $(1,1)$-tensor $\mathrm{R}_{i}^{~{}j}$ with itself, $\mathrm{R}_{i}^{~{}k}\mathrm{R}_{k}^{~{}j}$, which we denote for brevity by $\mathrm{Ric}^{2}$. We also denote its dual, the bilinear form $\mathrm{R}_{i}^{~{}k}\mathrm{R}_{kj}$, by $\mathrm{Ric}^{2}$. If $\mathrm{Ric}^{2}=0$ we say that $\mathfrak{g}$ has $2$-step nilpotent Ricci tensor. Note that $2$-step nilpotency of the Ricci tensor cannot occur for Riemannian metrics as it implies that the image of the Ricci endomorphism is totally null, i.e, light-like,

[TABLE]

where $\mathrm{g}_{kl}$ is the metric defined by the $\mathrm{ad}$-invariant scalar product $\langle.,.\rangle$. Moreover, $2$-step nilpotency of the Ricci endomorphism implies that its image is contained in its kernel. In addition, $\mathrm{Ric}^{2}=0$ implies the vanishing of the scalar curvature (see for example [2] for more details).

The following lemma holds for metric Lie algebras and is crucial for reducing the Bach tensor to the Ricci tensor and its square.

Lemma 3.1**.**

Let $\mathfrak{g}$ be a metric Lie algebra. Then its Ricci tensor satisfies

[TABLE]

where $K$ is the Killing form and $\mathrm{R}$ the curvature of $\mathfrak{g}$ defined by the metric $\mathrm{g}=\langle.,.\rangle$. Written in index notation, this is

[TABLE]

where $c_{ak}^{\ \ p}$ are the structure constants of the Lie algebra $\mathfrak{g}$.

Proof.

We fix a basis $(X_{1},\ldots,X_{n})$ of $\mathfrak{g}$, and use indices for the components of tensors in this basis, where $\mathrm{R}_{ij}$ denotes the components of $\mathrm{Ric}$. Then we compute

[TABLE]

by the $\mathrm{ad}$-invariance of $K$. This implies the first stated equality. Similarly, we obtain

[TABLE]

which gives the second equality. ∎

This lemma allows us to compute a formula for the Bach tensor that only involves the Ricci tensor:

Proposition 3.1**.**

The Bach tensor of a metric Lie algebra of dimension $n$ satisfies

[TABLE]

where $\rho$ is the scalar curvature of $\mathfrak{g}$ and $\mathrm{g}=\langle.,.\rangle$ is the metric.

Proof.

For a metric Lie algebra the Cotton tensor is $\mathrm{A}_{abc}=0$, so that a straightforward computation shows

[TABLE]

This allows to use Lemma 3.1 to replace contractions $\mathrm{R}^{ad}\mathrm{R}_{abcd}$ of the curvature tensor with $\mathrm{Ric}$ by $\mathrm{Ric}^{2}$-terms. ∎

This formula for the Bach tensor yields a proof of Theorem 1.2 from the introduction.

Proof of Theorem 1.2.

The proof is based on the following observation.

Lemma 3.2**.**

Let $A\in\mathrm{End}(\mathbb{R}^{n})$ be a linear map of $\mathbb{R}^{n}$ such that

[TABLE]

Then one of the following two cases occurs:

1. (1)

$A=\lambda\mathbf{1}+N$, with some $N\in\mathrm{End}(\mathbb{R}^{n})$ such that $N^{2}=0$. When $N\not=0$, then

[TABLE] 2. (2)

$A$* is diagonalisable with only two different eigenvalues $\lambda$ and $\mu$, in which case*

[TABLE]

Conversely, every $A$ as in (1) or (2) satisfies eq. 3.1.

Proof.

This lemma follows entirely from the Jordan normal form for $A$. If $A$ is considered as a linear map on $\mathbb{C}^{n}$, then squaring a Jordan block of size $3$ or larger shows that relation (3.1) is not satisfied, which implies that such blocks cannot occur.

Next assume that $A$ has at least one Jordan block $J_{\lambda}$ of size $2$. Then the equation for $A$ implies that $b=-2\lambda$ and $c=\lambda^{2}$, which shows that $\lambda$ is real. In this case any eigenvalue $\mu$ of $A$, then satisfies

[TABLE]

and hence $\lambda=\mu$. This is equivalent to $A=\lambda\mathbf{1}+N$ with $N\not=0$ and $N^{2}=0$.

Finally assume that $A$ is diagonalisable. Then eq. 3.1 for $A$ implies that $A$ can have at most two eigenvalues $\lambda$ and $\mu$, in which case $b=-\lambda-\mu$ and $c=\lambda\mu$. The case of $\lambda=\mu$ is contained in the first alternative.

For the converse it is straightforward to check that every $A$ as in (1) or (2) satisfies eq. 3.1. ∎

For the proof of Theorem 1.2 first we assume that $\mathfrak{g}$ is Bach-flat, use Proposition 3.1 and apply the lemma to the Ricci tensor $A=\mathrm{Ric}$. The Bach flatness then implies that

[TABLE]

Hence, by Lemma 3.2, we get that either $\mathrm{Ric}$ is diagonalisable with two different eigenvalues or $\mathrm{Ric}=A=\lambda\mathbf{1}+N$ with $N^{2}=0$ and with $\rho=\mathrm{tr}(A)=\lambda n$ and $\mathrm{tr}(A^{2})=n\lambda^{2}$. In the latter case, if $N=0$, the Ricci tensor is a multiple of the metric and hence Einstein, so we assume that $N\not=0$. The requirements on $b$ and $c$ from Lemma 3.2 then give that

[TABLE]

Since $n>2$ was assumed, this can only hold if $\lambda=0$. Hence $\mathrm{Ric}=N$ is $2$-step nilpotent.

In the case when $\mathrm{Ric}=A$ is diagonalisable with two eigenvalues $\lambda\not=\mu$ we have that $\rho=k\lambda+(n-k)\mu$, $\mathrm{tr}(A^{2})=k\lambda^{2}+(n-k)\mu^{2}$, where $k$, $1\leq k<n$, is the dimension of the eigenspace of $\lambda$ and $n-k$ the dimension of the eigen space of $\mu$. With $b$ and $c$ from Lemma 3.2 we get

[TABLE]

The left equation implies that

[TABLE]

This together with $\lambda+\mu\in\mathbb{R}$ implies that both eigenvalues are real. Also their eigenspaces $V_{\lambda}$ and $V_{\mu}$ are orthogonal to each other: since $\lambda\not=\mu$,

[TABLE]

shows that $\mathrm{g}(v_{\lambda},v_{\mu})=0$. Then the eigenspaces are also complementary, this implies that they are non degenerate with respect to $\mathrm{g}$. An important observation is that Equation 3.2 implies that the kernel of $A$ is either trivial or of dimension $1$. If the kernel of $A$ has dimension $1$, it is spanned by a non-degenerate vector. This proves the only if direction in Theorem 1.2.

For the converse recall that every semi-Riemannian Einstein manifold is Bach-flat. Moreover, if the Ricci tensor of $\mathfrak{g}$ is $2$-step nilpotent, the scalar curvature vanishes, and by Proposition 3.1 this implies that the Bach tensor vanishes. For the remaining case in Theorem 1.2 with eigenvalues $\lambda\not=0$ and $\mu=-\tfrac{n-k-1}{k-1}\lambda$ one checks that $b=-\frac{\rho}{(n-1)}$ and $c=\frac{1}{n}\left(\frac{\rho^{2}}{n-1}-\mathrm{tr}(\mathrm{Ric}^{2})\right)$ satisfy the conditions on $b$ and $c$ in Lemma 3.2, which then implies that $\mathfrak{g}$ is Bach flat by Proposition 3.1. ∎

Finally in this section, let us make a remark about algebraic properties of $\mathfrak{g}$ implying Bach flatness. It is well known that bi-invariant metrics on nilpotent Lie groups are Ricci-flat. Solvable Lie algebras in general are not Ricci flat, but from Cartan’s solvability criterion it follows that the Killing form vanishes on the derived Lie algebra. Hence, from Proposition 3.1 and noting that $2$-step nilpotent linear maps have vanishing trace, we obtain the following result (which we could not locate in the literature so far):

Corollary 3.2**.**

Let $\mathfrak{g}$ be a solvable metric Lie algebra. Then $\mathfrak{g}$ has two-step nilpotent Ricci tensor, vanishing scalar curvature, and is Bach flat.

4. Simple metric Lie algebras

For simple Lie groups, the space of bi-invariant metrics can be described explicitly.

Proposition 4.1**.**

Let $\mathfrak{g}$ be a real simple Lie algebra. Then either

1. (1)

The complexification $\mathfrak{g}^{\mathbb{C}}$ is simple and the space of $\mathrm{ad}$-invariant symmetric bilinear form is one-dimensional spanned by the Killing form of $\mathfrak{g}$, or 2. (2)

the complexification $\mathfrak{g}^{\mathbb{C}}$ is not simple and $\mathfrak{g}=\mathfrak{h}_{\mathbb{R}}$ is equal to a complex simple Lie algebra $\mathfrak{h}$ considered as real Lie algebra $\mathfrak{h}_{\mathbb{R}}$. In this case the space of $\mathrm{ad}$-invariant symmetric bilinear forms is two-dimensional and spanned by the real and imaginary part of the Killing form of $\mathfrak{h}$.

This proposition essentially follows from Schur’s lemma. It also yields a classification of $\mathrm{ad}$-invariant symmetric bilinear forms of semi-simple real Lie algebras. It was noted in [23], [21] and [1], and can be obtained from results in [8] and the fact that simple Lie algebras do not admit ad-invariant skew symmetric bilinear forms.

Clearly, when the complexification $\mathfrak{g}^{\mathbb{C}}$ is simple, any bi-invariant metric $\mathrm{g}$ is defined by the Killing form of $\mathfrak{g}$ and hence Einstein. The other case is treated in the following theorem.

Theorem 4.1**.**

Let $\mathfrak{g}$ be a real simple metric Lie algebra of dimension $n$ with $\mathrm{ad}$-invariant scalar product $\mathrm{g}$ and assume that $\mathfrak{g}^{\mathbb{C}}$ is not simple. Then $\mathfrak{g}=\mathfrak{h}_{\mathbb{R}}$ for a complex simple Lie algebra $\mathfrak{h}$ of dimension $m=\frac{n}{2}$ and $\mathrm{g}$ is of neutral signature $(m,m)$ given by

[TABLE]

where $K_{R}$ and $K_{I}$ are the bi-invarant symmetric bilinear forms corresponding to the real and imaginary part of the Killing form $K_{\mathfrak{h}}$ of $\mathfrak{h}$ and $\lambda$ and $\mu$ are real constants. The scalar curvature $\rho$ and Ricci and Bach tensors of $\mathrm{g}$ are given by

[TABLE]

Moreover, $\mathrm{g}$ is conformally Einstein if and only if

1. (1)

$\mu=0$, in which case $\mathrm{g}$ is Einstein, or when 2. (2)

$\lambda=0$* and $\mathfrak{h}=\mathfrak{sl}_{2}\mathbb{C}$, in which case $\mathrm{g}$ is conformally flat.*

Proof.

Let $\mathfrak{h}_{0}$ be the split real form of $\mathfrak{h}$ such that $\mathfrak{g}=\mathfrak{h}_{\mathbb{R}}=\mathfrak{h}_{0}\oplus\mathrm{i}\mathfrak{h}_{0}$ and let $X_{a}$, $a=1,\ldots,m$, be a basis of $\mathfrak{h}_{0}$ and of $\mathfrak{h}$. Denote by $K_{ab}=K_{\mathfrak{h}}(X_{a},X_{b})$ the matrix of the Killing form $K_{\mathfrak{h}}$ of $\mathfrak{h}$ in this basis. Then in the basis $X_{a},\mathrm{i}X_{b}$ of $\mathfrak{g}=\mathfrak{h}_{\mathbb{R}}$, the bilinear forms $K_{R}=\mathrm{Re}\circ K_{\mathfrak{h}}$ and $K_{I}=\mathrm{Im}\circ K_{\mathfrak{h}}$ considered as bilinear forms on $\mathfrak{g}=\mathfrak{h}_{\mathbb{R}}$ are of the form

[TABLE]

Note that $K_{R}=\tfrac{1}{2}K_{\mathfrak{g}}=-2\mathrm{Ric}$. The formulae for the scalar curvature and the Bach tensor are obtained from Proposition 3.1 by direct computation using that the metric and its inverse are

[TABLE]

where $K^{ab}$ is the inverse matrix of $K_{ab}$.

Now assume that $\mathfrak{g}$ is conformally Einstein and apply Corollary 3.1. Then $\mathfrak{g}$ is Bach flat only if $\lambda\mu=0$. In the case $\mu=0$ the metric is Einstein, so we assume $\mu\not=0$ and $\lambda=0$, i.e., $\langle.,.\rangle=\mu K_{I}$, which is not Einstein, but has vanishing scalar curvature. Moreover, since $\mathfrak{g}=\mathfrak{h}_{\mathbb{R}}$ and as there is no complex simple Lie algebra of dimension $2$, the dimension of $\mathfrak{g}$ is even but strictly greater than $4$. Hence, both conditions for the second obstruction in Corollary 3.1 to vanish are satisfied and we conclude that the Weyl tensor $\mathrm{C}$ has a non-trivial Weyl nullity ideal. Since $\mathrm{C}$ is $\mathrm{ad}$-invariant, its kernel is also $\mathrm{ad}$-invariant, and hence, with $\mathfrak{g}$ being simple, we have $\ker(\mathrm{C})=\mathfrak{g}$, i.e., $\mathrm{C}\equiv 0$. A computation of the Weyl tensor of the metric $\mu K_{I}$ on $\mathfrak{g}=\mathfrak{h}_{\mathbb{R}}$ yields

[TABLE]

for all $X,Y,Z,W\in\mathfrak{h}$. As $K_{\mathfrak{h}}$ is a complex bilinear form, this shows that $\mathrm{C}\equiv 0$ implies that

[TABLE]

If the rank of the complex Lie algebra $\mathfrak{h}$ is $\geq 2$, this gives a contradiction: in eq. 4.1, when taking $X$ and $Y=Z$ from a Cartan subalgebra of $\mathfrak{h}$ such that $K_{\mathfrak{h}}(Y,Y)\not=0$ and $K_{\mathfrak{h}}(X,Y)=0$, the left hand side vanishes (as the Cartan subalgebra is abelian) and thus

[TABLE]

for all $W\in\mathfrak{h}$, which contradicts the non-degeneracy of the Killing form.

When the rank of $\mathfrak{h}$ is $1$, i.e., when $\mathfrak{h}=\mathfrak{sl}_{2}\mathbb{C}$, this argument breaks down and it can be checked directly that equation eq. 4.1 is indeed satisfied. Taking the imaginary part of this equation, then shows that for $\mathfrak{h}=\mathfrak{sl}_{2}\mathbb{C}$ the metric $\mathrm{g}=\mu K_{I}$ on $\mathfrak{g}=\mathfrak{h}_{\mathbb{R}}$ has $\mathrm{C}=0$, i.e., is conformally flat. ∎

5. Metric Lie algebras via double extensions

5.1. Double extensions

Let $\mathfrak{h}$ be metric Lie algebra with $\mathrm{ad}$-invariant scalar product $\langle.,.\rangle_{\mathfrak{h}}$ and let $\mathfrak{s}$ be Lie algebra. Moreover, let $\delta:\mathfrak{s}\to\mathfrak{der}(\mathfrak{h})\cap\mathfrak{so}(\mathfrak{h})$ be a Lie algebra homomorphism into the skew symmetric derivations of $\mathfrak{h}$, that is, $\langle\delta_{S}(X),Y\rangle_{\mathfrak{h}}=-\langle X,\delta_{S}(Y)\rangle_{\mathfrak{h}}$ where we denote by $\delta_{S}$ the image of $S\in\mathfrak{s}$ under $\delta$. Also, for $X\in\mathfrak{h}$ we denote by $\delta_{\bullet}(X)$ the linear map from $\mathfrak{s}$ to $\mathfrak{h}$ that sends $S$ to $\delta_{S}(X)$.

In this setting the first step in defining a double extension of $\mathfrak{h}$ is to define the central extension of $\mathfrak{h}$ by $\mathfrak{s}^{*}$, where $\mathfrak{s}^{*}$ is the dual vector space to $\mathfrak{s}$, that is given by the cocycle in $Z^{2}(\mathfrak{h},\mathfrak{s}^{*})$ defined by

[TABLE]

We denote this central extension by $\mathfrak{s}^{*}\oplus_{\delta}\mathfrak{h}$. Recalling the definition of a central extension given by a cocycle, the Lie bracket of $\mathfrak{s}^{*}\oplus_{\delta}\mathfrak{h}$ is given by

[TABLE]

Next, we consider the adjoint representation $\mathrm{ad}$ of $\mathfrak{s}$ and its dual $\mathrm{ad}^{*}$, the co-adjoint representation of $\mathfrak{s}$ on $\mathfrak{s}^{*}$ given by $\mathrm{ad}^{*}_{S}(\sigma):=\sigma\circ\mathrm{ad}_{S}\in\mathfrak{s}^{*}$. Similarly, we denote by $\mathrm{ad}^{*}_{\bullet}(\sigma)$ the map that sends $S\in\mathfrak{s}$ to $\mathrm{ad}^{*}_{S}(\sigma)\in\mathfrak{s}^{*}$. This allows us to extend the map $\delta:\mathfrak{s}\to{\mathfrak{der}}(\mathfrak{h})$ to a map from $\mathfrak{s}$ to ${\mathfrak{der}}(\mathfrak{s}^{*}\oplus_{\delta}\mathfrak{h})$, which we also denote by $\delta$,

[TABLE]

Definition 5.1**.**

Let $\mathfrak{h}$ be metric Lie algebra with $\mathrm{ad}$-invariant scalar product $\langle.,.\rangle_{\mathfrak{h}}$, let $\mathfrak{s}$ be Lie algebra with an $\mathrm{ad}$-invariant bilinear form $\mathrm{b}$ and let $\delta:\mathfrak{s}\to\mathfrak{der}(\mathfrak{h})\cap\mathfrak{so}(\mathfrak{h})$ be a Lie algebra homomorphism into the skew derivations of $\mathfrak{h}$. Then the double extension $\mathfrak{g}$ of $\mathfrak{h}$ by $\mathfrak{s}$ and $\delta$, is the metric Lie algebra that is given by the semidirect sum of the central extension $\mathfrak{s}^{*}\oplus_{\delta}\mathfrak{h}$ with $\mathfrak{s}$ and the the map $\delta$ in eq. 5.1,

[TABLE]

together with the $\mathrm{ad}$-invariant inner product $\widetilde{\mathrm{g}}$ given by

[TABLE]

Recalling the definition of the semidirect sum $\rtimes_{\delta}$, the Lie bracket in $\mathfrak{g}$ is given in the splitting $\mathfrak{s}^{*}\oplus\mathfrak{h}\oplus\mathfrak{s}$ by

[TABLE]

or in terms of the adjoint representation

[TABLE]

for $\sigma\in\mathfrak{s}^{*}$, $X\in\mathfrak{h}$ and $S\in\mathfrak{s}$, and $\mathrm{ad}^{\mathfrak{h}}$ the adjoint representation of $\mathfrak{h}$. In particular, $\mathfrak{s}^{*}$ is an abelian ideal and $(\mathfrak{s}^{*})^{\perp}=\mathfrak{s}^{*}\oplus_{\delta}\mathfrak{h}$ is an ideal in $\mathfrak{g}$, and that $\mathfrak{s}$ is a subalgebra of $\mathfrak{g}$. A double extension admits several exact sequences of Lie algebras, for $\mathfrak{g}$

[TABLE]

and one for $(\mathfrak{s}^{*})^{\perp}$,

[TABLE]

The importance of double extensions stems from the remarkable structure result for indecomposable metric Lie algebras by Medina & Revoy [22] (see Theorem 1.1 in our Introduction), which states that every nonabelian, nonsimple, indecomposable metric Lie algebra is a double extension by a simple or a $1$-dimensional Lie algebra. An interesting algebraic fact that was already observed in [4] is the following:

Lemma 5.1**.**

Let $\mathrm{g}=(\mathfrak{s}^{*}\oplus_{\delta}\mathfrak{h})\rtimes_{\delta}\mathfrak{s}$ be a double extension by an abelian Lie algebra $\mathfrak{s}$. Then the metric Lie algebras $(\mathfrak{g},\langle.,.\rangle_{\mathrm{b}})$ and $(\mathfrak{g},\langle.,.\rangle_{0})$ are isomorphic as metric Lie algebras.

Proof.

Let $S_{1},\ldots,S_{r}$ be a basis of $\mathfrak{s}$, $\mathrm{b}_{ij}=\mathrm{b}(S_{i},S_{j})$, and $\sigma^{i}$ a dual basis to $S_{i}$. Then the vector space isomorphism $F:\mathfrak{g}\to\mathfrak{g}$ defined by

[TABLE]

is an isometry between $(\mathfrak{g},\langle.,.\rangle_{\mathfrak{g},\mathrm{b}})$ and $(\mathfrak{g},\langle.,.\rangle_{\mathfrak{g},0})$, that is, $F^{*}\langle.,.\rangle_{\mathfrak{g},\mathrm{b}}=\langle.,.\rangle_{\mathfrak{g},0}$. That $F$ is also a Lie algebra isomorphism can be easily checked using the assumption that $\mathfrak{s}$ is abelian. ∎

Remark 5.1**.**

The computation that is used to show that $F$ is a Lie algebra homomorphism breaks down when $\mathfrak{s}$ is not abelian. Indeed, let $c_{~{}ij}^{k}$ be the structure constants of $\mathfrak{s}$. The only terms that prevent $F$ from being a Lie algebra homomorphism are

[TABLE]

using eq. 5.2 and the $\mathrm{ad}^{\mathfrak{s}}$-invariance of $\mathrm{b}$, i.e., that $\mathrm{b}_{kl}c_{~{}ij}^{k}=\mathrm{b}_{ki}c_{~{}jl}^{k}=-\mathrm{b}_{kj}c_{~{}il}^{k}$, in the two last steps.

In the case of abelian $\mathfrak{s}$, the result in Lemma 5.1 allows us to assume without loss of generality that $\mathrm{b}=0$, in which case we denote the $\mathrm{ad}^{\mathfrak{g}}$-invariant inner product by $\langle.,.\rangle:=\langle.,.\rangle_{0}$.

Now, for a metric double extension, we will provide a formula for the Ricci tensor (equivalently its Killing form, see also [4]) and its square. Let $\mathfrak{g}=(\mathfrak{s}^{*}\oplus_{\delta}\mathfrak{h})\rtimes_{\delta}\mathfrak{s}$ be a metric double extension with invariant scalar product $\langle.,.\rangle$ and let $\mathrm{Ric}$ be its Ricci tensor. We identify $\mathrm{Ric}$ with $-\tfrac{1}{4}K$, where $K$ is the Killing form of $\mathfrak{g}$. Multiplying two matrices in (5.3) and taking their trace shows that

[TABLE]

and

[TABLE]

Here $K_{\mathfrak{h}}$ and $K_{\mathfrak{s}}$ are the Killing forms of $\mathfrak{h}$ and $\mathfrak{s}$, and $\mathrm{tr}$ denotes the trace of a linear map. For future reference we define $\eta\in\mathfrak{h}^{*}\otimes\mathfrak{s}^{*}$ by

[TABLE]

Here, $\mathrm{e}_{i}$ is a basis of $\mathfrak{h}$. For brevity, we will also write $\mathrm{tr}(\delta^{2})\in\otimes^{2}\mathfrak{s}^{*}$ for $\mathrm{tr}(\delta_{S}\circ\delta_{\widehat{S}})$ when $S,\widehat{S}\in\mathfrak{s}$. For the square of $\mathrm{Ric}$ this implies

[TABLE]

where $\eta(.,S)^{\sharp}\in\mathfrak{h}$ denotes the dualisation of the one-form $\eta(.,S)\in\mathfrak{h}^{*}$ with respect to $\langle.,.\rangle_{\mathfrak{h}}$, i.e., $\langle\eta(.,S)^{\sharp},X\rangle_{\mathfrak{h}}=\eta(X,S)$ for all $X\in\mathfrak{h}$. Moreover the scalar curvature of $\mathfrak{g}$ is given by

[TABLE]

These observations and Theorem 1.2 allow us to prove Corollary 1.1 in the introduction:

Proof of Corollary 1.1.

Note that $\mathfrak{s}^{*}\makebox[7.0pt]{\rule{6.0pt}{0.3pt}\rule{0.3pt}{5.0pt}}\,\mathrm{Ric}=0$ but $\mathfrak{s}^{*}\makebox[7.0pt]{\rule{6.0pt}{0.3pt}\rule{0.3pt}{5.0pt}}\,\mathrm{g}\not=0$. This already implies the first equivalence, that $\mathfrak{g}$ is Einstein if and only if it is Ricci flat.

For the second equivalence, that $\mathfrak{g}$ is Bach flat if and only if $\mathrm{Ric}^{2}=0$, we use Theorem 1.2, and we have to exclude the possibility of a diagonalisable Ricci tensor For the double extension however, the abelian ideal $\mathfrak{s}^{*}$ is always in the kernel of $\mathrm{Ric}$, which excludes the case of two nonzero eigen values, so according to Theorem 1.2 we are left with a one dimensional and non-degenerate kernel of $\mathrm{Ric}$. But this contradicts the fact that $\mathfrak{s}^{*}$ is null. Hence the only remaining possibility for Bach flatness in Theorem 1.2 is $\mathrm{Ric}^{2}=0$. ∎

For completeness we collect a few observations that are interesting, but not necessarily needed for our main results. Here, in the case where we assume $\mathfrak{s}=\mathbb{R}$, we fix an $0\not=S\in\mathfrak{s}$ and a $\delta=\delta_{S}$.

Proposition 5.1**.**

Let $\mathfrak{g}=(\mathfrak{s}^{*}\oplus_{\delta}\mathfrak{h})\rtimes_{\delta}\mathfrak{s}$ be a double extension.

1. (1)

If $\mathfrak{g}$ is solvable/nilpotent, then $\mathfrak{h}$ and $\mathfrak{s}$ are solvable/nilpotent. 2. (2)

If $\mathfrak{h}$ and $\mathfrak{s}$ are solvable, then $\mathfrak{g}$ is solvable. 3. (3)

If $\mathfrak{h}$ is nilpotent, $\mathfrak{s}=\mathbb{R}$ and $\delta$ is a nilpotent derivation, then $\mathfrak{g}$ is nilpotent.

Proof.

The proof is based on the maps in the two exact sequences (5.5) and (5.6). First assume that $\mathfrak{g}$ is solvable/nilpotent. Then $\mathfrak{s}$, as a homomorphic image under the projection $q$ in (5.5), is solvable/nilpotent, and the subalgebra $(\mathfrak{s}^{*})^{\perp}$ is solvable/nilpotent. Hence by the projection $r$ in (5.6), $\mathfrak{h}$ is solvable/nilpotent.

For the second point assume that $\mathfrak{s}$ and $\mathfrak{h}=(\mathfrak{s}^{*})^{\perp}/(\mathfrak{s}^{*})$ are solvable. Since $\mathfrak{s}^{*}$ is a central ideal, this implies that $(\mathfrak{s}^{*})^{\perp}$ is a solvable ideal in $\mathfrak{g}$. But $\mathfrak{g}/(\mathfrak{s}^{*})^{\perp}\simeq\mathfrak{s}$ is solvable and consequently $\mathfrak{g}$ is solvable.

For the third point, recall Engel’s Theorem that a Lie algebra is nilpotent if and only if all its adjoints $\mathrm{ad}_{X}$ are nilpotent linear maps. It is easy to check that the linear maps $\mathrm{ad}_{(\sigma,X,S)}$ in eq. 5.3 are nilpotent whenever $\mathfrak{h}$ and $\delta$ are nilpotent. ∎

A counter example to the implication “$\mathfrak{h}$ and $\mathfrak{s}$ nilpotent implies $\mathfrak{g}$ nilpotent” is given by the oscillator algebras (see Section 6.1), for which both $\mathfrak{h}$ and $\mathfrak{s}=\mathbb{R}$ are abelian, but $\mathfrak{g}$ is only solvable but not nilpotent. Here $(\mathfrak{s}^{*})^{\perp}$ is not central in $\mathfrak{g}$. However, if $\mathfrak{h}$ is nilpotent and $\mathfrak{s}=\mathbb{R}$, we can describe the Ricci tensor more precisely:

Proposition 5.2**.**

Let $\mathfrak{g}$ be a double extension with $\mathfrak{h}$ nilpotent, $\mathfrak{s}=\mathbb{R}$ and a derivation $\delta$. Then $\mathrm{Ric}=\mathrm{tr}(\delta^{2})$ and $\mathrm{Ric}^{2}=0$, and hence $\mathrm{B}=0$.

Proof.

The proof is based on the following observation:

Lemma 5.2**.**

Let $\mathfrak{h}$ be a nilpotent Lie algebra and $\delta$ a derivation of $\mathfrak{h}$. Then for all $X\in\mathfrak{h}$, the linear map $\delta\circ\mathrm{ad}_{X}$ is nilpotent and in particular $\mathrm{tr}(\delta\circ\mathrm{ad}_{X})=0$.

Proof.

Let $\mathfrak{h}^{k}=[\mathfrak{h},\mathfrak{h}^{k-1}]$ be the lower central series of $\mathfrak{h}$ and $\delta$ a derivation, then

[TABLE]

This can be used to show inductively that $\delta$ preserves $\mathfrak{h}^{k}$ and that the image of $(\delta\circ\mathrm{ad}_{X})^{k}$ is contained in $\mathfrak{h}^{k}$. Hence with $\mathfrak{h}$ nilpotent, the image of $(\delta\circ\mathrm{ad}_{X})^{k}$ eventually becomes zero, which means that $(\delta\circ\mathrm{ad}_{X})^{k}$ is a nilpotent linear map. As a consequence it is trace free. ∎

Then with $\mathfrak{h}$ being nilpotent we have that $K_{\mathfrak{h}}=0$ and the Lemma implies that $\eta=0$, and hence $\mathrm{Ric}=\mathrm{tr}(\delta^{2})$ and $\mathrm{Ric}^{2}=0$. This implies $\mathrm{B}=0$. ∎

5.2. Double extensions by simple Lie algebras

Here we will show that double extensions by simple Lie algebras cannot be conformally Einstein. The key fact we are going to use in the following is the algebraic version of the Karpelevich-Mostow Theorem:

Theorem 5.1** (Karpelevich [15], Mostow [24], see also [25, Corollary 1 in §6]).**

Let $f:\mathfrak{s}\to\widehat{\mathfrak{s}}$ be a homomorphism of real semisimple Lie algebras and let $\mathfrak{s}=\mathfrak{k}\oplus\mathfrak{p}$ be a Cartan decomposition of $\mathfrak{s}$. Then there is a Cartan decomposition of $\widehat{\mathfrak{s}}=\widehat{\mathfrak{k}}+\widehat{\mathfrak{p}}$ such that $f(\mathfrak{k})\subset\widehat{\mathfrak{k}}$ and $f(\mathfrak{p})\subset\widehat{\mathfrak{p}}$.

Recall (e.g. from [18, Chapter VI], [27, Chapter 4] or [25]) that a Cartan decomposition of a real semisimple Lie algebra $\mathfrak{s}$ is a decomposition $\mathfrak{s}=\mathfrak{k}\oplus\mathfrak{p}$ such that

1. (1)

$\mathfrak{k}$ is a subalgebra, $[\mathfrak{k},\mathfrak{p}]\subset\mathfrak{p}$ and $[\mathfrak{p},\mathfrak{p}]\subset\mathfrak{k}$, and 2. (2)

the Killing form of $\mathfrak{s}$ is negative definite on $\mathfrak{k}$ and positive definite on $\mathfrak{p}$, and $\mathfrak{k}$ and $\mathfrak{p}$ are orthogonal to each other.

A Cartan subalgebra $\mathfrak{t}$ of $\mathfrak{s}$ is a subalgebra such that $\mathfrak{t}^{\mathbb{C}}$ is a Cartan subalgebra of $\mathfrak{s}^{\mathbb{C}}$ (see [7, Sections 2.3.1 and 2.3.7] or [19]). Given a Cartan decomposition $\mathfrak{s}=\mathfrak{k}\oplus\mathfrak{p}$, there is a stable Cartan subalgebra $\mathfrak{t}\not=0$, i.e., a Cartan subalgebra such that $\mathfrak{t}=(\mathfrak{t}\cap\mathfrak{k})\oplus(\mathfrak{t}\cap\mathfrak{p})$. The dimension of a Cartan subalgebra is the rank of $\mathfrak{s}$. Using these facts, Theorem 5.1 implies the following statement that is useful for our purposes:

Corollary 5.1**.**

Let $\mathfrak{s}\subset\mathfrak{so}(t,s)$ be a subalgebra in $\mathfrak{so}(t,s)$ that is semisimple. Let $\mathfrak{s}=\mathfrak{k}\oplus\mathfrak{p}$ be a Cartan decomposition and $\mathfrak{t}$ a stable Cartan subalgebra. Assume that the Killing form of $\mathfrak{s}$ satisfies

[TABLE]

with some $\lambda\in\mathbb{R}$, where $\mathrm{tr}$ denotes the trace form in $\mathfrak{so}(t,s)$. Then $\lambda>0$.

Proof.

Let $\mathfrak{s}=\mathfrak{k}\oplus\mathfrak{p}$ be Cartan decomposition of $\mathfrak{s}$ and $\mathfrak{so}(t,s)=\widehat{\mathfrak{k}}\oplus\widehat{\mathfrak{p}}$ the associated Cartan decomposition of $\mathfrak{so}(t,s)$ with $\mathfrak{k}=\widehat{\mathfrak{k}}\cap\mathfrak{s}$ and $\mathfrak{p}=\widehat{\mathfrak{p}}\cap\mathfrak{s}$ and let $\mathfrak{t}$ and $\widehat{\mathfrak{t}}$ two stable Cartan subalgebras of $\mathfrak{s}$ and $\mathfrak{so}(t,s)$. Then, from the above properties of Cartan decompositions and Killing forms, it follows that both Killing forms $K_{\mathfrak{so}(t,s)}$ and $K_{\mathfrak{s}}$ are negative definite on $\mathfrak{t}\cap\mathfrak{k}=\widehat{\mathfrak{t}}\cap\widehat{\mathfrak{k}}$ and positive definite on $\mathfrak{t}\cap\mathfrak{p}=\widehat{\mathfrak{t}}\cap\widehat{\mathfrak{p}}$. Using the assumption and the relation between the Killing form of $\mathfrak{so}(t,s)$ and the trace form we get

[TABLE]

Taking $X,Y\in\mathfrak{t}\cap\mathfrak{k}$ or in $\mathfrak{t}\cap\mathfrak{p}$ implies that $\lambda>0$. ∎

Returning to double extensions by simple Lie algebras we get the following result:

Proposition 5.3**.**

Let $\mathfrak{g}=(\mathfrak{s}^{*}\oplus_{\delta}\mathfrak{h})\rtimes_{\delta}\mathfrak{s}$ be a double extension of a metric Lie algebra $\mathfrak{h}$ by a simple Lie algebra $\mathfrak{s}$. Then $\mathrm{Ric}|_{\mathfrak{s}\times\mathfrak{s}}\not=0$. In particular, $\mathfrak{g}$ cannot be Einstein.

Proof.

Since $\mathfrak{s}$ is simple, $\delta(\mathfrak{s})\subset\mathfrak{so}(t,s)$ is either trivial or simple. From the computation of the Ricci tensor in (5.7) we have seen that

[TABLE]

Assuming that this vanishes for all $S,T\in\mathfrak{s}$ gives a contradiction: if $\delta(\mathfrak{s})$ trivial, then $K_{\mathfrak{s}}=0$, which contradicts the simplicity of $\mathfrak{s}$, and if $\delta(\mathfrak{s})$ is simple it is in contradiction with Corollary 5.1 applied to $\delta(\mathfrak{s})\subset\mathfrak{so}(t,s)$.

Finally, assume that the double extension $\mathfrak{g}$ is Einstein. Then by Corollary 1.1 $\mathfrak{g}$ is Ricci-flat which contradicts $\mathrm{Ric}|_{\mathfrak{s}\times\mathfrak{s}}\not=0$. ∎

A version of this Theorem in the case that $\mathfrak{h}$ is abelian was stated in [4, Theorem 4.1].

Next, in order to analyse the second conformal to Einstein condition for double extensions by a simple Lie algebra $\sigma$, we describe ideals in such double extensions. For this we use the two projections $p$ and $q$ in the exact sequences (5.4) and (5.5). The projection $q$ is simply given by $q(\sigma,X,S)=S\in\mathfrak{s}$.

Lemma 5.3**.**

Let $(\mathfrak{g}=(\mathfrak{s}^{*}\oplus_{\delta}\mathfrak{h})\rtimes_{\delta}\mathfrak{s},\langle.,.\rangle_{\mathfrak{g},\mathrm{b}})$ be a metric double extension by a simple Lie algebra $\mathfrak{s}$, and let $\mathfrak{n}$ be an ideal in $\mathfrak{g}$. Then $\mathfrak{n}$ contains $\mathfrak{s}^{*}$ or $\mathfrak{n}\subset\mathfrak{s}^{*}\oplus_{\delta}\mathfrak{h}$.

Proof.

Let $\mathfrak{n}$ be an ideal in $\mathfrak{g}$. Then, as the projection $q$ in the sequence (5.5) is surjective, $q(\mathfrak{n})\subset\mathfrak{g}/(\mathfrak{s}^{*}\oplus\mathfrak{h})\simeq\mathfrak{s}$ is an ideal in $\mathfrak{s}$. Since $\mathfrak{s}$ is simple, this implies that $q(\mathfrak{n})$ is either trivial, in which case $\mathfrak{n}\subset\mathfrak{s}^{*}\oplus_{\delta}\mathfrak{h}$ and the lemma is proven, or isomorphic to $\mathfrak{s}$. In the latter case, for an arbitrary element in $\eta=(\sigma,X,S)\in\mathfrak{n}$ we get from eq. 5.2 that

[TABLE]

Since $\mathfrak{s}$ is simple, it is $\mathfrak{s}=[\mathfrak{s},\mathfrak{s}]$, and hence this shows that $\mathfrak{s}^{*}\subset\mathfrak{n}$. ∎

Lemma 5.4**.**

Let $(\mathfrak{g}=(\mathfrak{s}^{*}\oplus_{\delta}\mathfrak{h})\rtimes_{\delta}\mathfrak{s},\langle.,.\rangle_{\mathfrak{g},\mathrm{b}})$ be a metric double extension of $\mathfrak{h}$ by a simple Lie algebra $\mathfrak{s}$. Assume that $\mathfrak{g}$ has vanishing scalar curvature, so the Killing form of $\mathfrak{h}$ is trace free, and let $\mathfrak{n}$ be the Weyl nullity ideal. If $\mathfrak{n}$ contains $\mathfrak{s}^{*}$, then $\mathfrak{s}=\mathfrak{sl}_{2}\mathbb{R}$ or $\mathfrak{s}=\mathfrak{so}(3)$ and

[TABLE]

Proof.

Assume that $\mathfrak{s}^{*}\subset\mathfrak{n}$. Hence we have the condition $\sigma\makebox[7.0pt]{\rule{6.0pt}{0.3pt}\rule{0.3pt}{5.0pt}}\,\mathrm{C}=0$ for every $\sigma\in\mathfrak{s}^{*}$. Evaluating this for $X,Y,Z\in\mathfrak{s}$, and assuming that the scalar curvature of $\mathfrak{g}$ vanishes, we get

[TABLE]

where $K_{\mathfrak{s}}$ is Killing form of $\mathfrak{s}$. If the rank of $\mathfrak{s}$ is not $1$, i.e., if the dimension of a Cartan subalgebra $\mathfrak{t}$ is greater than $1$, we take linearly independent $X,Y$ in $\mathfrak{t}$ and get that $0=\mathrm{C}(X,Y,Z,\sigma)$ yields

[TABLE]

If $\mathfrak{t}$ is a stable Cartan subalgebra we can use Corollary 5.1 and the same argument as in the proof of Proposition 5.3 leads to a contradiction.

In the rank $1$ case, we have $\mathfrak{s}=\mathfrak{so}(3)$ or $\mathfrak{s}=\mathfrak{sl}_{2}\mathbb{R}$. The result is then a direct computation using that for these $\mathfrak{s}$ we have

[TABLE]

Moreover, $\mathrm{C}(H,Y,Z,\sigma)=0$ yields $\mathrm{Ric}(H,Z)=0$ for all $H\in\mathfrak{h}$ and $Z\in\mathfrak{g}$. ∎

These lemmas enable us to prove Theorem 1.4 in the introduction, which states that double extensions by simple Lie algebras cannot be conformally Einstein unless $\mathfrak{s}=\mathfrak{so}(3)$ or $\mathfrak{s}=\mathfrak{sl}_{2}\mathbb{R}$..

Proof of Theorem 1.4.

Let $\mathfrak{h}$ be a metric Lie algebra and $\mathfrak{g}=(\mathfrak{s}^{*}\oplus_{\delta}\mathfrak{h})\rtimes_{\delta}\mathfrak{s}$ a double extension with $\mathfrak{s}$ simple. If $\mathfrak{g}$ is conformally Einstein, by Theorem 1.2, the first condition $\mathrm{Ric}^{2}=0$ is satisfied, which implies that the scalar curvature vanishes. Moreover, by Proposition 5.3, $\mathfrak{g}$ is not Einstein, so we can use the vanishing of the second obstruction in Corollary 3.1. Let $\mathfrak{n}$ be the non trivial Weyl nullity ideal.

We can assume that the rank of $\mathfrak{s}$ is at least $2$. Then by Lemmas 5.3 and 5.4 we have that $\mathfrak{s}^{*}\not\subset\mathfrak{n}$ and hence $\mathfrak{n}\subset\mathfrak{s}^{*}\oplus_{\delta}\mathfrak{h}$. We consider the projection

[TABLE]

which is a Lie algebra homomorphism, and we denote $\mathfrak{n}_{0}:=\operatorname{pr}_{\mathfrak{h}}(\mathfrak{n})$. Since $\mathfrak{n}$ is an ideal for each $S\in\mathfrak{s}$ and $\sigma+H\in\mathfrak{n}$ we have $[\sigma+H,S]=-\mathrm{ad}_{S}^{*}(\sigma)-\delta_{S}(H)\in\mathfrak{n}$, and hence that $\mathfrak{n}_{0}$ is invariant under all derivations in the image of $\delta$, i.e., under $\delta(\mathfrak{s})$.

Using the vanishing of the scalar curvature again, we will evaluate the condition 2.5: For a nonzero element $N=\sigma+H\in\mathfrak{n}$, and any $X,Y,Z\in\mathfrak{g}$ we get

[TABLE]

By setting $Y=\widehat{\sigma}\in\mathfrak{s}^{*}$ we get

[TABLE]

for all $X,Z\in\mathfrak{g}$ and hence we get $(\mathfrak{n}_{0})\makebox[7.0pt]{\rule{6.0pt}{0.3pt}\rule{0.3pt}{5.0pt}}\,\mathrm{Ric}=0$. The above equation for the Weyl nullity ideal then simplifies to

[TABLE]

for all $X,Y,Z\in\mathfrak{g}$. If $X,Y,Z\in\mathfrak{s}$, then as $\mathfrak{s}$ is a subalgebra and $\mathfrak{s}\perp\mathfrak{h}$, this implies the same equation as in the proof of Lemma 5.4,

[TABLE]

for all $X,Y,Z\in\mathfrak{s}$ and for $\sigma\in\operatorname{pr}_{\mathfrak{s}^{*}}(\mathfrak{n})$, where $\operatorname{pr}_{\mathfrak{s}^{*}}:\mathfrak{s}^{*}\oplus_{\delta}\mathfrak{h}\to\mathfrak{s}^{*}$ is the projection onto $\mathfrak{s}^{*}$. If the projection of $\mathfrak{n}$ onto $\mathfrak{s}^{*}$ is equal to $\mathfrak{s}^{*}$, we get a contradiction in the same way as in the proofs of Lemma 5.4 and Proposition 5.3, so we may assume that $\operatorname{pr}_{\mathfrak{s}^{*}}(\mathfrak{n})\not=\mathfrak{s}^{*}$.

Now we consider $\mathfrak{n}^{\perp}$, the ideal in $\mathfrak{g}$ that is orthogonal to $\mathfrak{n}$ with respect to $\langle.,.\rangle$. As we have $\operatorname{pr}_{\mathfrak{s}^{*}}(\mathfrak{n})\not=\mathfrak{s}^{*}$, the ideal $\mathfrak{n}^{\perp}$ contains a non-trivial subspace $\mathfrak{s}\cap\mathfrak{n}^{\perp}$. Since $\mathfrak{s}$ is a subalgebra in $\mathfrak{g}$, the subspace $\mathfrak{s}\cap\mathfrak{n}^{\perp}$ is in fact an ideal in $\mathfrak{s}$ and hence equal to $\mathfrak{s}$, because of the simplicity of $\mathfrak{s}$. Now, since $\mathfrak{h}\perp\mathfrak{s}$, the inclusions $\mathfrak{s}\subset\mathfrak{n}^{\perp}$ and $\mathfrak{n}\subset\mathfrak{s}^{*}\oplus_{\delta}\mathfrak{h}$ imply that $\mathfrak{n}\subset\mathfrak{h}$. But since $\mathfrak{n}$ is an ideal, the bracket relation (5.2) then implies that $\delta_{S}(H)=0$ for all $S\in\mathfrak{s}$ and $H\in\mathfrak{n}_{0}=\mathfrak{n}$. With this information, and with $\sigma=0$, eq. 5.12 becomes

[TABLE]

for all $Y\in\mathfrak{h}$ and $S,T\in\mathfrak{s}$. Since $\langle.,.\rangle_{\mathfrak{h}}$ is non-degenerate, this implies that $\mathrm{Ric}(S,T)=0$ for all $S,T\in\mathfrak{s}$, which leads again to a contradiction as in the proof Proposition 5.3. ∎

The following example shows that in the case when the rank of $\mathfrak{s}$ is $1$ (see Lemma 5.4), the obstructions can vanish without the metric being conformally Einstein.

Example 5.1**.**

We will now present an example that is not governed by Theorem 1.4. Here, $\mathfrak{s}$ is of rank $1$, and we describe a double extension for which both obstructions vanish, i.e., which is Bach flat and has a nontrivial Weyl nullity ideal, but which however is not conformally Einstein.

Let $\mathfrak{s}=\mathfrak{so}(3)$, $\mathfrak{h}=\mathbb{R}^{3}$ be abelian with the Euclidean standard inner product, and $\delta=\frac{\sqrt{6}}{2}\,\mathrm{Id}_{\mathfrak{so}(3)}$ a map from $\mathfrak{s}$ to the derivations of the abelian Lie algebra $\mathbb{R}^{3}$ in $\mathfrak{so}(3)$. That is, $\delta_{S}=\frac{\sqrt{6}}{2}\,S$ for all $S\in\mathfrak{so}(3)$. Let $\mathfrak{g}=(\mathfrak{s}^{*}\oplus_{\delta}\mathbb{R}^{3})\rtimes_{\delta}\mathfrak{s}$ be the double extension of $\mathfrak{h}=\mathbb{R}^{3}$ by $\mathfrak{s}=\mathfrak{so}(3)$ and $\delta$ with the inner product $\langle.,.\rangle_{0}$, i.e., with $\mathrm{b}=0$. Let $E_{i}\in\mathbb{R}^{3}$, $i=1,2,3$, be the Cartesian standard basis of $\mathbb{R}^{3}$, and $S_{i}\in\mathfrak{s}$, $i=1,2,3$, be a basis of $\mathfrak{so}(3)$ such that

[TABLE]

where $(i,j,k)$ is an even permutation of $(1,2,3)$. Moreover, let $\sigma^{i}\in\mathfrak{s}^{*}$ be the dual basis to $S_{i}$, so $\sigma^{i}(S_{j})=\delta^{i}_{~{}j}$. Then $(\sigma^{1},\sigma^{2},\sigma^{3},E_{1},E_{2},E_{2},S_{1},S_{2},S_{3})$ forms a basis of $\mathfrak{g}$. The non-vanishing Lie brackets in $\mathfrak{g}$ are

[TABLE]

where $(i,j,k)$ again is an even permutation of $(1,2,3)$.

We are now going to show that for the simply connected metric Lie group

[TABLE]

both obstructions vanish but that $\mathfrak{g}$ as metric Lie algebra is not locally conformally Einstein, admitting no solution to eq. 2.3.

First we notice that the Schouten tensor of $G$ is given by

[TABLE]

where $\mathring{\mathrm{g}}$ and $\mathring{\mathsf{P}}$ are the round metric and its Schouten tensor on $\mathbb{S}^{3}\simeq\mathbf{SU}(2)$. In particular, $G$ is scalar flat and Bach flat so the first obstruction vanishes. Also the second obstruction vanishes, since a direct computation as in the proof of Lemma 5.3 using eq. 5.11 shows that $\mathfrak{s}^{*}$ is contained in the Weyl nullity ideal, $\mathfrak{s}^{*}\subset\mathfrak{n}$.

In regards to Equation 2.3, we are going to show that in fact $\mathfrak{n}=\mathfrak{s}^{*}$. Since $\mathfrak{s}^{*}\subset\mathfrak{n}$, it is enough to show that there is no nontrivial $V+S\in\mathbb{R}^{3}\oplus\mathfrak{s}$ that annihilates the Weyl tensor. For such $V+S$ and all $X,Y,Z\in\mathfrak{g}$ the Weyl tensor is

[TABLE]

Now choosing $X=S_{i}$ and $Y=S_{j}$ and $Z\in\mathbb{R}^{3}$, and recalling that $\left[S_{k},Z\right]\in\mathbb{R}^{3}$ and hence orthogonal to $\mathfrak{s}$, leads to the condition,

[TABLE]

where $(i,j,k)$ is an even permutation of $(1,2,3)$. But we also have that $\left[\mathfrak{s},\mathbb{R}^{3}\right]=\mathbb{R}^{3}$, which implies $V=0$ as $\langle.,.\rangle$ is non-degenerate on $\mathbb{R}^{3}$. Finally, choosing $X=E_{i}$ and $Y=E_{j}\in\mathbb{R}^{3}$ and $Z=S_{l}\in\mathfrak{s}$ we have

[TABLE]

where $(i,j,k)$ is an even permutation of $(1,2,3)$. Since $[\mathfrak{s}^{*},\mathfrak{s}]=\mathfrak{s}^{*}$, this implies $S=0$, and we can conclude that $\mathfrak{n}=\mathfrak{s}^{*}$.

Next we assume that there is a rescaling function $\varphi\in C^{\infty}(G)$ such that $\mathrm{e}^{2\varphi}\mathrm{g}$ is Einstein. Since $\mathfrak{n}=\mathfrak{s}^{*}$, the gradient $\nabla\varphi$ of $\varphi$ is tangent to $\mathfrak{n}=\mathfrak{s}^{*}$, i.e.,

[TABLE]

for some functions $a_{i}\in C^{\infty}(G)$, and where $\sigma^{i}\in\mathfrak{s}^{*}\subset\Gamma(TG)$. Hence, for the differential of $\varphi$ we have

[TABLE]

where now $\sigma^{i}\in\mathfrak{s}^{*}\subset\Gamma(T^{*}\mathbf{SU}(2))$. Note that the $\sigma^{i}$’s here are understood as sections of $T^{*}G$ as well as vectors in $\mathfrak{s}^{*}\subset\Gamma(TG)$, but also as $1$-forms on $\mathbb{S}^{3}$. Then $0={\mathrm{d}}^{2}\varphi={\mathrm{d}}a_{i}\wedge\sigma^{i}+a_{i}{\mathrm{d}}\sigma^{i}$ shows that

[TABLE]

This implies that the $a_{i}$’s are actually smooth functions on $\mathbb{S}^{3}=\mathbf{SU}(2)$ only. Hence $\varphi\in C^{\infty}(\mathbb{S}^{3})$, with $\Upsilon={\mathrm{d}}\varphi=a_{i}\sigma^{i}$, for $a_{i}\in C^{\infty}(\mathbb{S}^{3})$, is a solution to eq. 2.3, with the function $\lambda$ to be given by

[TABLE]

So $\Upsilon$ must satisfy the following equation,

[TABLE]

Note that, since $\mathrm{g}$ is not Einstein, $\Upsilon={\mathrm{d}}\varphi\not=0$.

We obtain the covariant derivatives of the $\sigma^{i}$’s, which are understood as vector fields on $G$ that are elements in the abelian ideal $\mathfrak{s}^{*}$ of $\mathfrak{g}$, from the bracket relations (5.13) as

[TABLE]

and all other covariant derivatives of $\sigma^{i}$ being zero. More concisely for $\nabla\sigma^{i}\in\Gamma(TG\otimes T^{*}G)$, we have

[TABLE]

and

[TABLE]

where $\mathring{\nabla}$ is the Levi-Civita connection of the round $3$-sphere and $(i,j,k)$ is an even permutation of $(1,2,3)$. Hence, with Equation 5.14, our crucial Equation 5.15 becomes

[TABLE]

which is an equation only on $\mathbb{S}^{3}$. With $\mathring{\mathsf{P}}=\tfrac{1}{2}\mathring{\mathrm{g}}$, this equation can be rewritten to a version of the conformally Einstein equation (2.3),

[TABLE]

on $\mathbb{S}^{3}$ with $\lambda=\frac{1}{4}$. This implies that $\varphi$ is in fact a local rescaling of the round metric $\mathring{\mathrm{g}}$  to another Einstein metric $\widehat{\mathrm{g}}=\mathrm{e}^{2\varphi}\mathring{\mathrm{g}}$ on $\mathbb{S}^{3}$. The round metric on the sphere however is locally conformally flat, and whence the metric $\widehat{\mathrm{g}}$ is a locally conformally flat Einstein metric. Consequently, $\widehat{\mathrm{g}}$ is a metric of constant curvature $\kappa$ with Schouten tensor $\widehat{\mathsf{P}}=\frac{\kappa}{2}\widehat{\mathrm{g}}=\frac{\kappa\mathrm{e}^{2\varphi}}{2}\mathring{\mathrm{g}}$. The transformation of the Schouten tensor in Equation 2.1 then yields

[TABLE]

This together with Equation 5.17 implies that

[TABLE]

Then, if $V$ is the metric dual of $\Upsilon={\mathrm{d}}\varphi$, i.e. $V$ is the the gradient of $\varphi$, we have

[TABLE]

On the other hand it holds that

[TABLE]

and therefore that

[TABLE]

When inserting $V$ into Equation 5.17, this yields

[TABLE]

Together with $\Upsilon\not=0$, this leads to a contradiction222Michael Eastwood showed us a more conceptual way of producing this contradiction, which uses the linearisation trick in Remark 2.1: For an arbitrary constant $c$, the PDE $\mathring{\nabla}\Upsilon-\Upsilon^{2}=c\mathring{\mathrm{g}}$ on the unit sphere $\mathbb{S}^{3}$ turns into the linear PDE $\mathring{\nabla}\mathring{\nabla}\sigma+c\sigma\mathring{\mathrm{g}}=0$ when substituting $\varphi=-\log(\sigma)$. Prolonging this equation yields a connection whose parallel sections correspond to solutions of the linear PDE. Computing its curvature shows that the connection has nontrivial parallel sections only when $c=1$. and shows that $\mathrm{g}$ is not conformally Einstein, even though it is Bach flat and has a nontrivial Weyl nullity ideal.

5.3. Double extensions by $\mathbb{R}$

When $\mathfrak{s}=\mathbb{R}$, we fix a nonzero vector $S$ in $\mathfrak{s}$ and its dual $\sigma\in\mathfrak{s}^{*}$, i.e., $\sigma(S)=1$, and denote by $\delta$ the corresponding derivation of $\mathfrak{h}$. Moreover, $\mathfrak{s}^{*}$ is contained in the centre of the double extension $\mathfrak{g}=(\mathfrak{s}^{*}\oplus_{\delta}\mathfrak{h})\rtimes_{\delta}\mathfrak{s}$. Lemma 5.1 implies we can assume that $\mathrm{b}=0$ without loss of generality in the definition of the double extension $\mathfrak{g}$. We will be able to make further simplifications because of the following result:

Theorem 5.2** ([10], see also [12]).**

Let $\mathfrak{h}$ be a Lie algebra, let $\delta$ and $\widehat{\delta}$ be two derivations of $\mathfrak{h}$ in $\mathfrak{so}(\mathfrak{h})$, and let $\mathfrak{g}$ and $\widehat{\mathfrak{g}}$ be the double extensions of $\mathfrak{h}$ by $\delta$ and $\widehat{\delta}$ respectively. Then there is an isomorphism $F:\mathfrak{g}\to\widehat{\mathfrak{g}}$ of metric Lie algebras if and only if there is a $\lambda\in\mathbb{R}$, an $X\in\mathfrak{h}$, and an isomorphism $f:\mathfrak{h}\to\mathfrak{h}$ of metric Lie algebras, such that

[TABLE]

In the following we will work with a basis $\mathrm{e}_{-},\mathrm{e}_{1},\ldots,\mathrm{e}_{n-2},\mathrm{e}_{+}$ of $\mathfrak{g}$ of the form $\mathrm{e}_{-}=\sigma$, $\mathrm{e}_{a}$, $a=1,\ldots,n-2$, a basis of $\mathfrak{h}$ such that $\mathrm{h}_{ab}=\langle\mathrm{e}_{a},\mathrm{e}_{b}\rangle_{\mathfrak{h}}$ are constants, and $\mathrm{e}_{+}=S$. We will use the following index convention: greek indices will run from $-,1,\ldots,n-2,+$ whereas latin indices run from only from $1$ to $n-2$. In this basis the metric satisfies

[TABLE]

The Ricci tensor of a double extension by $\mathbb{R}$, when written in this basis is given by

[TABLE]

where $(K_{\mathfrak{h}})_{ab}$ is the Killing form of $\mathfrak{h}$, and $\eta_{a}$ are

[TABLE]

Hence, the square of the Ricci tensor is

[TABLE]

Assume that the first obstruction vanishes, i.e., that the manifold is Bach flat. By Corollary 1.1 this yields $\mathrm{Ric}^{2}=\mathrm{R}_{\mu\kappa}\mathrm{R}^{\kappa}_{~{}\nu}$, which implies that the scalar curvature vanishes, we will now evaluate the second condition. That is, we will consider the existence of a nonzero element $0\not=V=V^{\mu}\mathrm{e}_{\mu}=V^{-}\mathrm{e}_{-}+V^{a}\mathrm{e}_{a}+V^{+}\mathrm{e}_{+}$ in the Weyl nullity ideal,

[TABLE]

where $\mathrm{C}$ is the Weyl tensor and where we use that $\mathrm{R}_{\alpha\beta\gamma-}=0$ and $\mathrm{R}_{\alpha-}=0$. This already provides us with a first solution to $V\makebox[7.0pt]{\rule{6.0pt}{0.3pt}\rule{0.3pt}{5.0pt}}\,\mathrm{C}$ in a special situation:

Proposition 5.4**.**

If $\mathfrak{g}$ is a double extension of $\mathfrak{h}$ by $\mathbb{R}$ and $\delta$ such that $\mathrm{Ric}=\mathrm{tr}(\delta^{2})$, that is if $K_{\mathfrak{h}}=0$ and $\eta=0$, then $\mathrm{e}_{-}\makebox[7.0pt]{\rule{6.0pt}{0.3pt}\rule{0.3pt}{5.0pt}}\,\mathrm{C}=0$, so $\mathrm{e}_{-}$ is in the Weyl nullity ideal.

As an example, Proposition 5.2 shows that the assumptions in this proposition are satisfied when $\mathfrak{h}$ is nilpotent.

Hence, from now on we assume that there is at least one index pair $(c,\beta)$ such that $\mathrm{R}_{c\beta}\not=0$. Setting $\alpha=-$ in eq. 5.20 we get

[TABLE]

which, when evaluated for $\gamma=+$ and $\gamma=a$, implies the conditions

[TABLE]

for all $a=1,\ldots n-2,\ \beta=1,\ldots,n-2,+$. This first equation implies that the $\mathfrak{h}$-component $\widehat{V}$ of $V$ is in the kernel of $\mathrm{Ric}$. Hence, since we have assumed that there is an index pair $(c,\beta)$ such that $\mathrm{R}_{c\beta}\not=0$, we have that $V^{+}=0$ and therefore that $V$ is in the kernel of $\mathrm{Ric}$. Now we evaluate the above equations for $\alpha=\gamma=+$ and $\beta=b$,

[TABLE]

This shows that the linear map $A:\mathfrak{h}\to\mathfrak{h}$, defined by $A(X)=\mathrm{R}(X,\mathrm{e}_{+})\mathrm{e}_{+}$, which in fact is given by $-\tfrac{1}{4}\delta^{2}$, satisfies

[TABLE]

as $\mathrm{R}_{++}=\mathrm{tr}(A)=-\tfrac{1}{4}\mathrm{tr}(\delta^{2})$. This means we have

[TABLE]

Next, we look at $\alpha=+$, $\beta=b$ and $\gamma=c$ and use $V^{d}\mathrm{R}_{d+}=0$ from above, so

[TABLE]

which can again be rewritten as

[TABLE]

Here $K_{\mathfrak{h}}^{\sharp}$ is the endomorphism obtained by metric dual of the Killing form, or

[TABLE]

Note that the equation for $\alpha=a$, $\beta=b$ and $\gamma=+$ follows from this by the Jacobi identity

[TABLE]

yielding

[TABLE]

where $\mathrm{ad}_{\delta(V)}$ is understood to be dualised with $\mathrm{h}$. With this we arrive at a reformulation of the two obstructions for double extension by $\mathbb{R}$.

Proposition 5.5**.**

Let $\mathfrak{h}$ be a metric Lie algebra of dimension $m$ with Killing form $K_{\mathfrak{h}}$, a derivation $\delta$, and corresponding $1$-form $\eta(X)=\mathrm{tr}(\delta\circ\mathrm{ad}_{X})$. Let $\mathfrak{g}=(\mathbb{R}^{*}\oplus_{\delta}\mathfrak{h})\rtimes_{\delta}\mathbb{R}$ be the metric Lie algebra obtained by the double extension of $\mathfrak{h}$ by $\mathbb{R}$ and $\delta$. If $\mathfrak{g}$ is conformally Einstein, then it holds that:

1. (a)

$\mathrm{Ric}^{2}=0$, i.e., $K^{2}_{\mathfrak{h}}=0$, $\eta^{\sharp}\makebox[7.0pt]{\rule{6.0pt}{0.3pt}\rule{0.3pt}{5.0pt}}\,K_{\mathfrak{h}}=0$, and $\eta^{\sharp}$ is a null vector in $\mathfrak{h}$. 2. (b)

If $\dim(\mathfrak{g})>4$ and $\mathfrak{g}$ is not Ricci-flat (and hence not Einstein), then either $K_{\mathfrak{h}}=0$ and $\eta=0$ (in which case it is $\mathrm{e}_{-}\makebox[7.0pt]{\rule{6.0pt}{0.3pt}\rule{0.3pt}{5.0pt}}\,\mathrm{C}=0$), or there is $V=V^{-}\mathrm{e}_{-}+\widehat{V}\in\mathbb{R}^{*}\oplus\mathfrak{h}$ such that

[TABLE]

where $\mathrm{ad}$ and $[.,.]$ are those of $\mathfrak{h}$.

Note that (5.26) is just

[TABLE]

We will use these equations in the next section, where we will also deal with the cases $n\leq 4$.

6. The oscillator algebras and related double extensions

6.1. The oscillator algebras

Let $\mathbb{R}^{t,s}$ be a semi-Euclidean vector space of dimension $\ell=t+s$ and signature $(t,s)$, and denote by $\langle.,.\rangle$ the semi-Euclidean standard inner product. We want to doubly extend the abelian metric Lie algebra $\mathbb{R}^{t,s}$ by $\mathbb{R}$ and a linear map $\Phi\in\mathfrak{so}(t,s)$. If $\Phi$ has a kernel, the double extension $(\mathbb{R}\oplus_{\Phi}\mathbb{R}^{t,s})\ltimes_{\Phi}\mathbb{R}$ is isomorphic as metric Lie algebra to the direct sum of the kernel of $\Phi$ with $(\mathbb{R}\oplus_{\Phi}V)\ltimes_{\Phi}\mathbb{R})$, where $V$ is the image of $\Phi$, and hence decomposable. So from now on we assume that $\Phi\in\mathfrak{so}(t,s)$ is invertible, which implies that $\ell$ is even. In this situation, the oscillator algebra $\mathfrak{osc}_{\Phi}(t,s)$ of dimension $m=\ell+2$ and signature $(t+1,s+1)$ is then defined as the double extension of $\mathbb{R}^{t,s}$ by $\mathbb{R}$ and by $\Phi$, i.e.,

[TABLE]

We set $\mathfrak{osc}_{\Phi}(\ell)=\mathfrak{osc}_{\Phi}(0,\ell)\simeq\mathfrak{osc}_{\Phi}(\ell,0)$. The adjoint of $\mathfrak{osc}_{\Phi}(t,s)$ is given by

[TABLE]

and the $\mathrm{ad}$-invariant scalar product is $\langle.,.\rangle=2r\rho+\langle.,.\rangle_{t,s}$ and hence of signature $(t+1,s+1)$. Here we work in a basis $\mathrm{e}_{0},\mathrm{e}_{1},\ldots,\mathrm{e}_{\ell},\mathrm{e}_{\ell+1}$ with $\mathrm{e}_{0}=(1,0,0)$, $\mathrm{e}_{\ell+1}=(0,0,1)$ and $\mathrm{e}_{i}$, $i=1,\ldots,\ell$ a basis of $\mathfrak{h}$. We denote by $e_{\mu}^{*}$ the (algebraically) dual basis.

Remark 6.1**.**

As the only non abelian Lie algebra of dimension $2$ does not admit an $\mathrm{ad}$-invariant scalar product, every indecomposable double extension of dimension $\leq 4$ is isomorphic to $\mathfrak{osc}_{J}(2)$ with $J\in\mathfrak{so}(2)$.

The centre of $\mathfrak{osc}_{\Phi}(t,s)$ is $\mathbb{R}^{*}=\mathbb{R}\cdot\mathrm{e}_{0}$ and the derived Lie algebra is equal to $\mathbb{R}^{*}\oplus\mathbb{R}^{t,s}$, which is isomorphic to the Heisenberg algebra and hence nilpotent. Therefore:

Lemma 6.1**.**

The oscillator algebras $\mathfrak{osc}_{\Phi}(t,s)$ are solvable. Their Killing form is given by $K((\rho,X,r),(\sigma,Y,s))=rs\ \mathrm{tr}(\Phi^{2})$, i.e., $K=\mathrm{tr}(\Phi^{2})\,(\mathrm{e}_{\ell+1}^{*})^{2}$.

By Corollary 3.2 and Proposition 5.4 this implies that the oscillator algebras satisfy both conformally Einstein conditions:

Proposition 6.1**.**

The oscillator algebras are Bach flat and satisfy $V\makebox[7.0pt]{\rule{6.0pt}{0.3pt}\rule{0.3pt}{5.0pt}}\,C=0$ for $V=\mathrm{e}_{0}=(1,0,0)\in\mathbb{R}^{*}$ a central element.

For the Proof of Theorem 1.5, that the oscillator algebras are conformally Einstein, it remains to check that the vector field $\mathrm{e}_{0}$ on $G$ indeed satisfies the conditions in (A) before Definition 1.1: as $\mathrm{e}_{0}$ is a parallel vector field on the corresponding simply connected metric Lie group $G$, it is a gradient vector field, and because of the formula for the Killing form, $\mathrm{Ric}$ and $\mathsf{P}$, its metric dual satisfies equation eq. 2.3. Hence the oscillator algebras are conformally Einstein in the sense of Definition 1.1. $\square$

In the next section we will consider double extensions of the oscillator algebras, for which we will need to know their derivations.

Lemma 6.2**.**

Every derivation $\delta\in\mathfrak{so}(\mathfrak{osc}_{\Phi}(t,s))$ of $\mathfrak{osc}_{\Phi}(t,s)$ is of the form

[TABLE]

and the corresponding $1$-form $\eta$ is given by

[TABLE]

i.e., $\eta=\mathrm{tr}(\Psi\circ\Phi)\mathrm{e}_{\ell+1}^{*}$ and hence $\eta^{\sharp}=\mathrm{tr}(\Psi\circ\Phi)\mathrm{e}_{0}$.

Proof.

Since $\mathbb{R}^{*}$ is the center of $\mathfrak{osc}_{\Phi}(t,s)$, it is left invariant under $\delta$. This and the condition that $\delta$ is skew with respect to $\mathrm{h}$, imply that

[TABLE]

The conditions that $\delta$ is a derivation means that

[TABLE]

Multiplying the corresponding matrices yields the equations

[TABLE]

and hence $a=0$ and $[\Phi,\Psi]=0$. The formula for $\eta$ is derived by direct calculation. ∎

6.2. Double extensions of the oscillator algebras

First we use Theorem 5.2 to simplify the derivation we are using for the double extension of the oscillator algebra:

Lemma 6.3**.**

Let $\mathfrak{h}=\mathfrak{osc}_{\Phi}(t,s)$ be the oscillator algebra given by $\Phi\in\mathfrak{so}(t,s)$. Moreover, let $\Psi\in\mathfrak{so}(t,s)$ and $\delta=(\Psi,0)$ and $\widehat{\delta}=(\Psi,u)$ with $u\in\mathbb{R}^{t,s}$ two derivations of $\mathfrak{h}$ (with the notations as in Lemma 6.2). Then the double extensions of $\mathfrak{h}$ by $\delta$ and $\widehat{\delta}$ are isomorphic metric Lie algebras.

Proof.

This follows from Theorem 5.2 with $f=\mathrm{Id}$, $\lambda=1$ and $X=(0,\Psi^{-1}(u),0)^{\top}\in\mathfrak{h}$. ∎

Hence, without loss of generality, we can assume that $\delta=(\Psi,0)$. Let $\mathfrak{g}_{\Psi,\Phi}$ be the metric Lie algebra which is a $1$-dimensional double extension of an oscillator algebra $\mathfrak{h}=\mathfrak{osc}_{\Phi}(t,s)$ by the derivation $\delta=(\Psi,0)$, that is

[TABLE]

The ad-invariant scalar product $\langle.,.\rangle$ is of signature $(t+2,s+2)$ and $\mathfrak{g}_{\Psi,\Phi}$ is of dimension $n=m+2=\ell+4=t+s+4>4$. Since $\delta(\mathrm{e}_{0})=0$, the centre of $\mathfrak{g}_{\Psi,\Phi}$ is spanned by $\mathrm{e}_{-}$ as defined in the previous section and by $\mathrm{e}_{0}=(1,0,0)\in\mathfrak{osc}_{\Phi}(t,s)$. They correspond to two parallel vector fields on $\mathfrak{g}_{\Psi,\Phi}$. Moreover Lemma 6.2, or the fact that double extensions of oscillator algebras by $\mathbb{R}$ are solvable, imply:

Lemma 6.4**.**

The metric Lie algebras $\mathfrak{g}_{\Psi,\Phi}$ have $2$-step nilpotent Ricci tensor and hence are Bach flat.

Analysing the second obstruction by using Proposition 5.5 gives the following result:

Theorem 6.1**.**

Let $\mathfrak{g}_{\Psi,\Phi}$ be the metric Lie algebra that is obtained from an oscillator algebra $\mathfrak{osc}_{\Phi}(t,s)$ by $1$-dimensional double extension by the derivation $(\Psi,v)$. If $\mathfrak{g}_{\Psi,\Phi}$ is conformally Einstein, then either $\mathrm{tr}(\Psi\circ\Phi)=0$ and $\mathrm{tr}(\Phi^{2})=0$, or there is a nonzero vector in $W\in\mathbb{R}^{t,s}$ such that

[TABLE]

for $\ell=t+s$. Moreover, let $B$ be the trace form of $\mathfrak{so}(t,s)$, which is non-degenerate. Then the matrices $\Phi$ and $\Psi$ are either a multiple of each other or span a plane in $\mathfrak{so}(t,s)$ that is degenerate with respect to $B$.

Proof.

We assume that at least one of $\mathrm{tr}(\Phi^{2})=0$ or $\mathrm{tr}(\Psi\circ\Phi)$ is not zero, so one of $K_{\mathfrak{h}}$ or $\eta$ is not zero for $\mathfrak{h}=\mathfrak{osc}_{\Phi}(t,s)$. Then, by the virtue of Proposition 5.5, conformally Einstein implies the equations (5.23–5.26) for $\widehat{V}=\sum_{i=0}^{\ell+1}V^{i}\mathrm{e}_{i}$. Since $\eta=\mathrm{tr}(\Psi\circ\Phi)\mathrm{e}_{\ell+1}^{*}$, equation 5.26 evaluated for the pair $\mathrm{e}_{0}$ and $\mathrm{e}_{\ell+1}$ and 5.23 imply that

[TABLE]

and hence $V^{\ell+1}=0$. Then $\widehat{V}=V^{0}\mathrm{e}_{0}+W$ with $W=\sum_{i=1}^{\ell}V^{i}\mathrm{e}_{i}\in\mathbb{R}^{t,s}\subset\mathfrak{osc}_{\Phi}(t,s)$. Equation (5.24) gives two more equations:

[TABLE]

In the same way equation (5.25) yields equations,

[TABLE]

Finally, for the oscillator algebras equation (5.26) reduces to

[TABLE]

The only pair of arguments for which this equation gives a condition is $\mathrm{e}_{\ell+1}$ and $X\in\mathbb{R}^{t,s}$, for which we obtain

[TABLE]

for all $X\in\mathbb{R}^{t,s}$. This yields the equation

[TABLE]

Equations (6.2) and (6.4) can be concisely written as

[TABLE]

This system has a non-trivial solutions space if and only if

[TABLE]

Now observe that $[\Psi,\Phi]=0$ implies that $(\Psi\circ\Phi)^{2}W=\Psi^{2}\Phi^{2}W$, so that equations (6.1), (6.3) and (6.5) ensure in a remarkable way that $\det(B(\Psi,\Phi))=0$, so that equations (6.2) and (6.4) have indeed a non-trivial solution $(V^{-},V^{0})$. ∎

Now recall that the trace form $B$ of $\mathfrak{so}(\ell)$ is negative definite. Moreover, the indecomposability of $\mathfrak{g}_{\Psi,\Phi}$ implies that $\Psi$ and $\Phi$ are not multiples of each other (see [4, Proposition 7.1] and [16]). Therefore, Theorem 6.1 yields the following conclusion:

Corollary 6.1**.**

If the metric Lie algebra $\mathfrak{g}_{\Psi,\Phi}$ is of signature $(2,\ell+2)$ and conformally Einstein, then $\Psi$ is a multiple of $\Phi$. In particular, if $\mathfrak{g}_{\Psi,\Phi}$ is indecomposable, it is not conformally Einstein.

6.3. The remaining case in signature $(2,n-2)$

In [4] it was shown that every indecomposable, nonsimple metric Lie algebra $\mathfrak{g}$ in signature $(2,n-2)$ is isomorphic to one of the following cases:

1. (1)

$\mathfrak{g}=\mathfrak{osc}_{\Phi}(1,n-3)$, which is conformally Einstein by our Theorem 1.5; 2. (2)

to a double extension of an oscillator algebra, i.e.,

[TABLE]

we have seen in the previous section that they are not conformally Einstein whenever they are indecomposable; 3. (3)

or to a double extension of the direct sum of $\mathfrak{osc}_{\Phi}(n-5)$ with $\mathbb{R}$, i.e.,

[TABLE]

with $\Phi\in\mathfrak{so}(n-5)$ with $\Phi\not=0$.

In the remainder we will show that the last case (3) is not conformally Einstein. Since $\mathfrak{h}=\mathfrak{osc}_{\Phi}(n-5)\oplus\mathbb{R}$ is solvable, its double extension is solvable and hence Bach flat, so we will focus on the second criterion, the non trivial Weyl nullity ideal, to show that $\mathfrak{g}$ is not conformally Einstein. These are the conditions in (b) of Proposition 5.5. For this we fix a basis $\mathrm{e}_{0},\ldots,\mathrm{e}_{\ell+2}$ of $\mathfrak{h}$, where $\mathrm{e}_{0},\ldots,\mathrm{e}_{\ell+1}$ is a basis of the oscillator algebra (as in the previous section), $\ell=n-5$, and $\mathrm{e}_{\ell+2}$ spans the central $\mathbb{R}$ direction. From Proposition 5.5 we get the existence of a $\widehat{V}=V^{0}\mathrm{e}_{0}+W+V^{\ell+1}\mathrm{e}_{\ell+1}+V^{\ell+2}\mathrm{e}_{\ell+2}$ with $W\in\mathbb{R}^{n-5}$, that satisfies equations (5.23–5.26).

The Killing form of $\mathfrak{h}$ is again given by $K_{\mathfrak{h}}=tr(\Phi^{2})\,(\mathrm{e}_{\ell+1}^{*})^{2}$. Since $\mathrm{e}_{0}$ and $\mathrm{e}_{\ell+2}$ are in the centre of $\mathfrak{h}$ and since $\mathrm{e}_{0}^{\flat}=\mathrm{e}_{\ell+1}^{*}$, $\mathrm{e}_{\ell+1}^{\flat}=\mathrm{e}_{0}^{*}$ and $\mathrm{e}_{\ell+2}^{\flat}=-\mathrm{e}_{\ell+2}^{*}$, equation (5.26) applied to the pairs $(\mathrm{e}_{0},\mathrm{e}_{\ell+1})$ and $(\mathrm{e}_{\ell+2},\mathrm{e}_{\ell+1})$ gives that

[TABLE]

Since $\Phi\in\mathfrak{so}(n-5)$, the vanishing of $\mathrm{tr}(\Phi^{2})$ would imply $\Phi=0$ which is excluded in this case. Hence, as for the oscillator algebra we have that $\widehat{V}=V^{0}\mathrm{e}_{0}+W$ with $W\in\mathbb{R}^{n-5}$.

Next we have to determine the derivations of a Lie algebra of the form

[TABLE]

As for the oscillator algebras, one can show [12] that the derivations of $\mathfrak{h}$ are of the form

[TABLE]

with $\Psi\in\mathfrak{so}(t,s)$, $c\in\mathbb{R}$, $u\in\mathbb{R}^{t,s}$ and $\left[\Phi,\Psi\right]=0$. Moreover, again using Theorem 5.2, one can show that every double extension of $\mathfrak{h}$ by $\mathbb{R}$ and $\delta$ with $c\not=0$ is isomorphic to a double extension by

[TABLE]

and hence we can assume this without loss of generality. As $\mathfrak{g}$ is indecomposable this implies that $\Phi$ and $\Phi$ are not a multiple of each other. The proof for these statements can be found in [16, 17] or [12], see also [4, Theorem 7.1].

From now on the proof that $\mathfrak{g}$ is not conformally Einstein proceeds with the derivation of equations (5.24–6.5) completely analogous to the proof of Theorem 6.1 and Corollary 6.1, for details see [12]. This yields the following conclusion, which gives a proof of Theorem 1.6:

Theorem 6.2**.**

Let $\mathfrak{g}=(\mathbb{R}^{*}\oplus_{\delta}\mathfrak{h})\rtimes_{\delta}\mathbb{R}$ be an indecomposable metric Lie algebra that is given by a double extension of $\mathfrak{h}=\mathfrak{osc}_{\Phi}(n-5)\oplus\mathbb{R}$ by a derivation $\delta$. Then $\mathfrak{g}$ is not conformally Einstein.