Energy in Fourth-Order Gravity

Abstract

In this paper we make a detailed analysis of conservation principles in the context of a family of fourth-order gravitational theories generated via a quadratic Lagrangian. In particular, we focus on the associated notion of energy and start a program related to its study. We also exhibit examples of solutions which provide intuitions about this notion of energy which allows us to interpret it, and introduce several study cases where its analysis seems tractable. Finally, positive energy theorems are presented in restricted situations.

Appendix

1.1 Geometric Conventions

1.1.1 Curvature Conventions

To avoid any ambiguity let us pinpoint the curvature conventions we follow in this text: the curvature tensor is defined as:

$$\begin{aligned} R(X,Y)Z=\nabla _X\nabla _YZ - \nabla _Y\nabla _XZ - \nabla _{[X,Y]}Z, \end{aligned}$$

with its components ordered as follows:

$$\begin{aligned} R^{i}_{jkl}&=dx^{i}(R(\partial _k,\partial _l)\partial _j),\\&=\partial _k\Gamma ^{i}_{lj}-\partial _l\Gamma ^{i}_{kj} + \Gamma ^{i}_{ku}\Gamma ^u_{jl}- \Gamma ^{i}_{lu}\Gamma ^u_{jk}.\\ \end{aligned}$$

From this we get the canonical Ricci, Einstein, and scalar tensors:

$$\begin{aligned} \textrm{Ric}_{ij}\doteq R^l_{ilj}. \end{aligned}$$

1.1.2 Differential Forms

In this section we will establish our conventions concerning differential forms and operations related to them. First of all, given a k-form $\omega \in \Omega ^k(V)$ on a d-dimensional manifold V and a local coordinate system $\{x^i\}_{i=1}^d$, we fix $\omega _{i_1\cdots i_k}\doteq \omega (\partial _{i_1},\cdots ,\partial _{i_k})$ as its components relative to the basis $\{dx^{j_1}\wedge \cdots dx^{j_k}\}_{j_1<\cdots <j_k}$, and therefore we may locally write

$$\begin{aligned} \omega =\sum _{i_1<\cdots <i_{k}}\omega _{i_1\cdots i_k}dx^{i_1}\wedge \cdots dx^{i_k}=\frac{1}{k!}\omega _{i_1\cdots i_k}dx^{i_1}\wedge \cdots dx^{i_k}, \end{aligned}$$

where in the last equality the summation is not restricted to $i_1<\cdots <i_k$. In this setting we have several well-known operations. To start with, given a semi-Riemannian manifold $(V^d,g)$ we have an induced semi-Riemannian metric $g^{(k)}$ on each space $\Omega ^k(V)$ which is given by

$$\begin{aligned} g^{(k)}(\alpha ,\beta )\doteq \sum _{i_1<\cdots <i_k}\alpha _{i_1\cdots i_k}\beta ^{i_1\cdots i_k}=\frac{1}{k!}\alpha _{i_1\cdots i_k}\beta ^{i_1\cdots i_k}, \text { for any } \alpha ,\beta \in \Omega ^k(V) \end{aligned}$$

where above $\beta ^{i_1\cdots i_k}\doteq g^{i_1j_1}\cdots g^{i_kj_k}\beta _{j_1\cdots j_k}$. It is not difficult to see that if $\{e_1,\cdots ,e_d\}$ is a g-orthonormal basis for $T_pV$ at a point $p\in V$, and if $\{e^1,\cdots ,$ $e^d\}$ stands for its dual basis, then $\{e^i_{1}\wedge \cdots \wedge e^{i_k}\}_{i_1<\cdots <i_k}$ is a $g^{(k)}$-orthonormal basis for the fibre of $\Omega ^k(V)$ over p (see, for instance, [2, Proposition 7.2.11]). In particular, it holds that

$$\begin{aligned} g^{(k)}(e^i_{1}\wedge \cdots \wedge e^{i_k},e^i_{1}\wedge \cdots \wedge e^{i_k})=c_{i_1}\cdots c_{i_k} \end{aligned}$$

where above $c_{i_{j}}\doteq g(e_{i_j},e_{i_j})=\pm 1$.

Also, in this setting, we introduce the Hodge-star operator, which is defined pointwise as a linear operator on each fibre. For this one needs to introduce a volume form. Thus, let us again consider an orientable semi-Riemannian manifold $(V^d,g)$ and denote the associated Riemannian volume form by $dV_g$, which locally reads as $dV_g=\sqrt{|\textrm{det}(g)|}dx^1\wedge \cdots \wedge dx^d$. Given our vector bundle $\Omega ^k(V)\xrightarrow []{\pi } V$, one defines a linear operator

$$\begin{aligned} \star _{g}:\Omega ^{k}(V)\mapsto \Omega ^{d-k}(V) \end{aligned}$$

by the requirement that for all $\alpha ,\beta \in \Omega ^{k}(V)$ it holds that

$$\begin{aligned} \beta \wedge \star _{g}\alpha =g^{(k)}(\beta ,\star _g \alpha )dV_g. \end{aligned}$$

Above, we add the subscript g on $\star$ to highlight the dependence on this operator on the choice of metric g. This can be seen to be well-defined and, in fact, in local (oriented) coordinates one can see that the action of $\star$ on $\alpha \in \Omega ^k(V)$ is given by (see, for instance, [2, Proposition 7.2.12 and Example 7.2.14(D)]):

$$\begin{aligned} \star _g\alpha&=\sum _{\underset{j_{k+1}<\cdots<j_d}{j_1<\cdots<j_k}}\alpha ^{j_1\cdots j_k}\epsilon _{j_1\cdots j_k j_{k+1}\cdots j_d}\sqrt{|\textrm{det}(g)|}dx^{j_{k+1}}\wedge \cdots \wedge dx^{j_d},\\ (\star _g\alpha )_{j_{k+1}\cdots j_{d}}&=\sum _{j_1<\cdots <j_k}\alpha ^{j_1\cdots j_k}\epsilon _{j_1\cdots j_k j_{k+1}\cdots j_d}\sqrt{|\textrm{det}(g)|}, \end{aligned}$$

where above $\epsilon _{j_1\cdots j_d}$ stands for the antisymmetric Levi-Civita symbol and the sums which appeal to Einstein summation convention are understood as over all possible values of the indices. Sometimes, we shall write $\mu _{j_1\cdots j_d}=\epsilon _{j_1\cdots j_d}$ $\sqrt{|\textrm{det}(g)|}$ which is defined via the relation

$$\begin{aligned} dV_g=\mu _{j_1\cdots j_d}dx^{j_1}\wedge \cdots \wedge dx^{j_d} \text { (no summation) }. \end{aligned}$$

Let us now recall that, given $\alpha \in \Omega ^k(V)$, the exterior differential $d:\Omega ^k(V)\mapsto \Omega ^{k+1}(V)$ is characterised by its local action given by

$$\begin{aligned} d\alpha =\sum _{i_1<\cdots <i_{k}}\partial _{i}\alpha _{i\cdots i_{k+1}}dx^{i}\wedge dx^{i_1}\wedge \cdots \wedge dx^{k}. \end{aligned}$$

(72)

In particular, given a 1-form $\alpha \in \Omega ^1(V)$, we see that

$$\begin{aligned} d\alpha&=\sum _{j=1}^d\partial _{i}\alpha _{j}dx^{i}\wedge dx^{j}=\partial _{i}\alpha _{j}dx^{i}\wedge dx^{j}=\sum _{i<j}(\partial _i\alpha _j-\partial _j\alpha _i)dx^i\wedge dx^j,\\ d\alpha _{ij}&=\partial _i\alpha _j-\partial _j\alpha _i. \end{aligned}$$

Now, given a semi-Riemannian manifold $(V^d,g)$ we can consider the formal adjoint $d^{*}:\Omega ^{k+1}(V)\mapsto \Omega ^{k}(V)$, which we shall denote by $\delta _{g}$ and where we shall highlight its dependence on g, which arises through the canonical choice of $dV_g$ as the volume form. One can thus globally write

$$\begin{aligned} \delta _g\alpha =(-1)^{dk+1+\textrm{Ind}(g)}\star _gd\star _g\alpha \text { for all } \alpha \in \Omega ^{k+1}(V), \end{aligned}$$

where $\textrm{Ind}(g)$ denotes the index of the semi-Riemannian metric g. For instance, if g is Lorentzian, we have $\textrm{Ind}(g)=1$. One can then compute that, in local coordinates, the following formula holds:

$$\begin{aligned} \delta _g\alpha _{i_1\cdots i_k}=-\nabla ^i\alpha _{ii_1\cdots i_k}, \end{aligned}$$

where above $\nabla$ stands for the Riemannian connection associated to g.

Also, given $X\in \Gamma (TV)$, let us introduce the interior product $X\lrcorner :\Omega ^k(V)\mapsto \Omega ^{k-1}(TV)$ on any manifold $V^d$, which is defined by

$$\begin{aligned}&X\lrcorner \alpha (X_1,\cdots ,X_{k-1})\doteq \alpha (X,X_1,\cdots ,X_{k-1}) \text { for any } \alpha \in \Omega (V)\\&\quad \text { and } \forall \,\, X_1,\cdots ,X_{k-1}\in \Gamma (TV). \end{aligned}$$

An important formula linking the exterior derivative, the interior product, and the Lie derivative is given by Cartan’s famous magic formula:

$$\begin{aligned} \pounds _X\alpha =d(X\lrcorner \alpha ) + X\lrcorner (d\alpha ) \,\, \forall \,\, \alpha \in \Omega ^k(V) \text { and } X\in \Gamma (TV). \end{aligned}$$

The above formula plays an important role in the application of Stokes’ theorem to operators in divergence form. Since in the core of this paper we will be interested in certain flux formalae which are derived in this manner within the Lorentzian setting, let us below briefly highlight a few differences with the more usual Riemannian setting.

Let $(V^{n+1}=M^n\times \mathbb {R}, \bar{g})$ be a Lorentzian manifold, parametrise the $\mathbb {R}$ factor with a coordinate t. Assume $\partial _t$ is time-like and denote by $g_t$ the induced Riemannian metric on each $M_t\doteq M\times \{t\}$. Assume furthermore that $(M^n,g_t)$ are orientable Riemannian manifolds and denote by $dV_{g_t}$ their corresponding volume forms. With all this, we have a natural orientation for V: if at $p\in M$ $\{e_1,\cdots ,e_n\}$ denotes a positive basis for $T_pM$, then $\{\partial _t,e_1,\cdots ,e_n\}$ denotes a positive basis for $T_{(p,t)}V$. Thus, if $\{x^i\}_{i=1}^n$ is a positively oriented coordinate system for M, then $dV_{\bar{g}}=\sqrt{|\textrm{det}(\bar{g})|}dt\wedge dx^1\wedge \cdots \wedge dx^n$ denotes our volume form.

Let $\Omega \subset M$ be a compact subset with smooth boundary and define the subset $\mathcal {C}_T=\Omega \times [0,T]$. This subset is Stokes regular, in the sense that it is regular enough so as to apply Stokes’ theorem over it. Now, let $X\in \Gamma (TV)$ and from the formula

$$\begin{aligned} L_XdV_{\bar{g}}=\textrm{div}_{\bar{g}}XdV_{\bar{g}}=d(X\lrcorner dV_{\bar{g}}) \end{aligned}$$

one gets that

$$\begin{aligned} \int _{\mathcal {C}_T}\textrm{div}_{\bar{g}}XdV_{\bar{g}}&=\int _{\partial \mathcal {C}_T}\mathcal {J}^{*}(X\lrcorner dV_{\bar{g}}) \end{aligned}$$

(73)

where above $\mathcal {J}^{*}:\partial \mathcal {C}_T\mapsto \mathcal {C}_T$ denotes the inclusion. We can split $\partial \mathcal {C}_{T}=\Omega _0\cup \Omega _T\cup L$, where $\Omega _0=\Omega \times \{0\}$, $\Omega _T=\Omega \times \{T\}$ and $L=\partial \Omega \times [0,T]$. On each of these hypersurfaces we denote the inclusion into V by $\mathcal {J}^{*}_0,\mathcal {J}^{*}_T$ and $\mathcal {J}^{*}_L$, respectively. Now, let n denote the future-pointing unit normal to each t-constant hypersurface $M_t=M\times \{t\}$. Then, writing $X=-\bar{g}(X,n)n + X^{\top }$, we find

$$\begin{aligned} \mathcal {J}^{*}_0(X\lrcorner dV_{\bar{g}})=\mathcal {J}^{*}_0(dV_{\bar{g}}(-\bar{g}(X,n)n + X^{\top },\cdot ))=-\bar{g}(X,n)\mathcal {J}^{*}_0(dV_{\bar{g}}(n,\cdot )). \end{aligned}$$

Notice that in (73) $\Omega _0$ is oriented with its Stokes induced orientation, where the outward-pointing unit normal corresponds to $-n$ and thus $\{e_1,\cdots ,e_n\}$ is a positive basis for $\Omega _0$ at p iff $\{-n,e_1,\cdots ,e_n\}$ is positive for $\mathcal {C}_T$. This implies that the induced Stokes orientation for $\Omega _0$ is actually opposite to its intrinsic orientation. On the other hand, we see that in the case of $\Omega _T$ these two orientations agree. All this implies that

$$\begin{aligned} \mathcal {J}^{*}_0(X\lrcorner dV_{\bar{g}})=-\bar{g}(X,n)dV_{\bar{g}_0} \, ,\, \mathcal {J}^{*}_T(X\lrcorner dV_{\bar{g}})=-\bar{g}(X,n)dV_{\bar{g}_T} \end{aligned}$$

and, using intrinsic orientations for $\Omega$,

$$\begin{aligned} \int _{\mathcal {C}_T}\textrm{div}_{\bar{g}}XdV_{\bar{g}}&=-\int _{\Omega _0}\mathcal {J}^{*}_0(X\lrcorner dV_{\bar{g}}) + \int _{\Omega _T}\mathcal {J}^{*}_T(X\lrcorner dV_{\bar{g}}) + \int _{L}\mathcal {J}^{*}_L(X\lrcorner dV_{\bar{g}}),\nonumber \\&=\int _{\Omega _0}\bar{g}(X,n)dV_{\bar{g}_0} - \int _{\Omega _T}\bar{g}(X,n)dV_{\bar{g}_T} + \int _{L}\mathcal {J}^{*}_L(X\lrcorner dV_{\bar{g}}), \end{aligned}$$

(74)

where ${J}^{*}_L(X\lrcorner dV_{\bar{g}})=g_t(X,\nu )\nu \lrcorner dV_{\bar{g}}=g_t(X,\nu )dL$, with $\nu$ the outward pointing unit normal vector field to L, which is to be understood with its induced Stokes orientation.

1.1.3 Extrinsic Geometry

In this section we shall quickly fix our conventions for the extrinsic curvature. Thus, let $M^n\hookrightarrow (V^{n+1},\bar{g})$ be an immersed hypersurface in a time-oriented Lorentzian manifold. We define the second fundamental form of M as

$$\begin{aligned} \mathbb{I}\mathbb{I}:\Gamma (TM)\times \Gamma (TM)&\mapsto \Gamma (TM^{\perp })\\ (X,Y)&\mapsto (\bar{\nabla }_{\bar{X}}\bar{Y})^{\perp } \end{aligned}$$

where $\bar{X}$ and $\bar{Y}$ denote arbitrary extensions (respectively) of X and Y to V, $\bar{\nabla }$ denotes the Riemannian connection associated to $\bar{g}$, and $TM^{\perp }$ denotes the normal bundle of M. Associated to the second fundamental form, we have the extrinsic curvature, here denoted by $K\in \Gamma (T^0_2M)$, which we define with respect to the future-pointing unit normal to M. Thus, K is defined by

$$\begin{aligned} K(X,Y)\doteq \bar{g}(\mathbb{I}\mathbb{I}(\bar{X},\bar{Y}),n)=\bar{g}(\bar{\nabla }_{\bar{X}}\bar{Y},n), \,\, \forall \, X,Y\in \Gamma (TM). \end{aligned}$$

Also, we define $\tau \doteq \textrm{tr}_gK$ as the (not normalised) mean curvature of the immersion, and therefore we find that

$$\begin{aligned} \tau =-\textrm{div}_{\bar{g}}n. \end{aligned}$$

Finally, let us notice that if $V^{n+1}=M^n\times \mathbb {R}$, with $\mathbb {R}$ parametrised by a coordinate t and the time orientation given by $\partial _t$, then defining the associated lapse N and shift X by

$$\begin{aligned} N=-\bar{g}(\partial _t,n),\,\, X=\partial _t-Nn. \end{aligned}$$

we may write $n=\frac{1}{N}(\partial _t-X)$. This implies that

$$\begin{aligned} -K(U,W)&=\bar{g}(\bar{U},\bar{\nabla }_{\bar{W}}n) =\frac{1}{N}\left( \bar{g}(\bar{U},\bar{\nabla }_{\bar{W}}\partial _t) - \bar{g}(\bar{U},\bar{\nabla }_{\bar{W}}X)\right) ,\\&=\frac{1}{N}\left( \partial _t(\bar{g}(\bar{U},\bar{W})) - \bar{g}(\bar{\nabla }_{\partial _t}\bar{U},\bar{W}) + \bar{g}(\bar{U},[\bar{W},\partial _t]) - g(U,\nabla _{W}X)\right) \end{aligned}$$

Therefore,

$$\begin{aligned} -2NK(U,W) =\partial _t(\bar{g}(\bar{U},\bar{W})) - \bar{g}(\bar{U},[\partial _t,\bar{W}]) - \bar{g}(\bar{W},[\partial _t,\bar{U}]) - \pounds _{X}g(U,W). \end{aligned}$$

That is,

$$\begin{aligned} K(U,W)=-\frac{1}{2N}\left( \partial _t(\bar{g}(\bar{U},\bar{W})) - \bar{g}(\bar{U},[\partial _t,\bar{W}]) - \bar{g}(\bar{W},[\partial _t,\bar{U}]) - \pounds _{X}g(U,W)\right) \end{aligned}$$

Notice that locally this reduces to a simple expression given by

$$\begin{aligned} K_{ij}=-\frac{1}{2N}\left( \partial _t\bar{g}_{ij} - \pounds _{X}g_{ij}\right) , \end{aligned}$$

from which we sometimes write $K=-\frac{1}{2N}\left( \partial _tg - \pounds _{X}g \right)$.

1.2 Proof of Proposition 4.1

Since the sketch presented has already dealt with the $2 \alpha + \beta = 0$ case, we will assume in the following that $\chi := 2 \alpha + \beta \ne 0$. Employing the same Maple procedure as presented in Fig. 1, we can see (see Fig. 3) that, as announced in the sketch of the proof: $M= -m -\frac{\Lambda }{3}r^3 +C_1 r^{f(\alpha ,\beta )} + C_2 r^{g(\alpha ,\beta )}$, with

$$\begin{aligned} f&= \frac{6 \alpha +3 \beta + \sqrt{100 \alpha ^2 + 84 \alpha \beta + 17 \beta ^2}}{2\left( 2 \alpha + \beta \right) } \\ g&= \frac{6 \alpha +3 \beta - \sqrt{100 \alpha ^2 + 84 \alpha \beta + 17 \beta ^2}}{2\left( 2 \alpha + \beta \right) } \end{aligned}$$

and

$$\begin{aligned} \frac{\left( 2 \alpha + \beta \right) ^2 r }{24 \left( \beta + 4 \alpha \right) }A_{22}&= C_1\left( h^+_1 r^{t_1^+(\alpha , \beta ) }+h^+_2r^{t_2^+(\alpha , \beta ) }+h^+_3 r^{t_3^+(\alpha , \beta ) } + h^+_4 r^{t_4^+(\alpha , \beta ) } \right) \\&\quad + C_2\left( h^-_1 r^{t_1^-(\alpha , \beta ) }+h^-_2 r^{t_2^-(\alpha , \beta ) }+h^-_3 r^{t_3^-(\alpha , \beta ) }+ h^-_4 r^{t_4^-(\alpha , \beta ) } \right) , \end{aligned}$$

with

$$\begin{aligned} t_1^-&=\frac{18 \alpha +9 \beta - \sqrt{100 \alpha ^2 + 84 \alpha \beta + 17 \beta ^2}}{4 \alpha + 2 \beta }\nonumber \\ t_2^-&=\frac{10 \alpha +5 \beta - \sqrt{100 \alpha ^2 + 84 \alpha \beta + 17 \beta ^2}}{4 \alpha + 2 \beta }\nonumber \\ t_3^-&=\frac{6 \alpha +3 \beta - \sqrt{100 \alpha ^2 + 84 \alpha \beta + 17 \beta ^2}}{4 \alpha + 2 \beta }\nonumber \\ t_4^-&=\frac{6 \alpha +3\beta - \sqrt{100 \alpha ^2 + 84 \alpha \beta + 17 \beta ^2}}{2 \alpha + \beta }\nonumber \\ t_1^+&=\frac{18 \alpha +9 \beta +\sqrt{100 \alpha ^2 + 84 \alpha \beta + 17 \beta ^2}}{4 \alpha + 2 \beta }\nonumber \\ t_2^+&=\frac{10 \alpha +5 \beta + \sqrt{100 \alpha ^2 + 84 \alpha \beta + 17 \beta ^2}}{4 \alpha + 2 \beta }\nonumber \\ t_3^+&=\frac{6 \alpha +3 \beta +\sqrt{100 \alpha ^2 + 84 \alpha \beta + 17 \beta ^2}}{4 \alpha + 2 \beta }\nonumber \\ t_4^+&=\frac{6 \alpha +3\beta +\sqrt{100 \alpha ^2 + 84 \alpha \beta + 17 \beta ^2}}{2 \alpha + \beta }. \end{aligned}$$

(75)

The $4 \alpha + \beta =0$ case will be treated separately as a special case.

Fig. 3

Necessary conditions on M

Remark 7.1

We must point out that $100 \alpha ^2 + 84 \alpha \beta + 17 \beta ^2$ is not necessarily positive. We will consider first that $100 \alpha ^2 + 84 \alpha \beta + 17 \beta ^2 \ge 0$, before explaining that the situation is highly similar in the opposite case.

Since $100 \alpha ^2 + 84 \alpha \beta + 17 \beta ^2 = 25 \left( 2 \alpha + \beta \right) ^2 -8 \beta \left( 2 \alpha + \beta \right) = 25 \chi ^2 - 8 \beta \chi$, we will favour working with $(\chi , \beta )$ instead of $(\alpha , \beta )$. We will thus write:

$$\begin{aligned} f&=\frac{3}{2}+\frac{\sqrt{25 -8 \frac{\beta }{\chi } } }{2} \nonumber \\ g&= \frac{3}{2}-\frac{\sqrt{25 -8 \frac{\beta }{\chi } } }{2}, \end{aligned}$$

(76)

and

$$\begin{aligned} t_1^-&=\frac{9}{2} -\frac{\sqrt{25 -8 \frac{\beta }{\chi } } }{2} \nonumber \\ t_2^-&=\frac{5}{2} -\frac{\sqrt{25 -8 \frac{\beta }{\chi } } }{2}\nonumber \\ t_3^-&=\frac{3}{2} -\frac{\sqrt{25 -8 \frac{\beta }{\chi } } }{2}\nonumber \\ t_4^-&=3 -\sqrt{25 -8 \frac{\beta }{\chi } }\nonumber \\ t_1^+&=\frac{9}{2} +\frac{\sqrt{25 -8 \frac{\beta }{\chi } } }{2}\nonumber \\ t_2^+&=\frac{5}{2} +\frac{\sqrt{25 -8 \frac{\beta }{\chi } } }{2}\nonumber \\ t_3^+&=\frac{3}{2} +\frac{\sqrt{25 -8 \frac{\beta }{\chi } } }{2}\nonumber \\ t_4^+&=3 +\sqrt{25 -8 \frac{\beta }{\chi } }. \end{aligned}$$

(77)

This of course works under the assumption that $\chi$ is positive. However, if $\chi \le 0$, $\frac{ \sqrt{ 100 \alpha ^2 + 84 \alpha \beta + 17 \beta ^2} }{\chi } = \textrm{sg}(\chi ) \sqrt{25 -8 \frac{\beta }{\chi }}$. Up to exchanging f and g, and the $t_i^+$ and the $t_i^-$, the coefficients remain the same.

Using these formulas, one can deduce that there exist a finite number of $[\alpha , \beta ] \in \mathbb {R}\mathbb {P}^1$ for which one of the $t_i^\pm$ equals another $t_j^\pm$. We present those values in the table, Fig. 4 and detail a few representative cases:

• $t_1^- =t_2^-, t_3^-$ clearly has no solution.

• $t_1^-=t_4^-$ is equivalent to $\frac{9}{2}-\frac{\sqrt{25 -8 \frac{\beta }{\chi } } }{2}=3 -\sqrt{25 -8 \frac{\beta }{\chi } }$ which is rephrased as $\sqrt{25 -8 \frac{\beta }{\chi } }=-3$. There are then no solutions.

• $t_1^-=t_1^+$ if and only if $\sqrt{25 -8 \frac{\beta }{\chi } }=0$, i.e. $\frac{\beta }{\chi }=\frac{25}{8}.$

• $t_1^-=t_2^+$ if and only if $\frac{9}{2} -\frac{\sqrt{25 -8 \frac{\beta }{\chi } } }{2}=\frac{5}{2} +\frac{\sqrt{25 -8 \frac{\beta }{\chi } } }{2}$, i.e. $\sqrt{25 -8 \frac{\beta }{\chi } }=2$, which means that: $\frac{\beta }{\chi }=\frac{21}{8}$.

All the other combinations fall into one of these configurations (obviously no solution, no solution because of negative squareroot, solution with null squareroot, solution with positive squareroot).

Fig. 4

Values of $\frac{\beta }{\chi }$ for which $t_i^\pm =t_j^\pm$

Outside of those specific values, the $\left( r^{t_i^\pm } \right)$ form a free family. Thus for the metric to be A flat one must have:

$$\begin{aligned} \left( -\left( \beta + 4 \alpha \right) \sqrt{ 100 \alpha ^2+ 84 \alpha \beta + 17 \beta ^2} + \left( 2 \alpha + \beta \right) \left( 7 \beta + 20 \alpha \right) \right) C_2=0. \end{aligned}$$

(78)

This equation is obtained by looking at the $r^{\frac{10 \alpha +5 \beta - \sqrt{100 \alpha ^2 + 84 \alpha \beta + 17 \beta ^2}}{4 \alpha + 2 \beta }}$ term (last term of the first line in formula (2) in Fig. 3). We will rephrase (78) in terms of $\beta$ and $\chi$:

$$\begin{aligned} C_2 \left( - \left( 2-\frac{\beta }{\chi } \right) \sqrt{ 25 - 8 \frac{\beta }{\chi } } +10- 3 \frac{\beta }{\chi } \right) =0, \end{aligned}$$

which implies that either $C_2=0$ or $\frac{\beta }{\chi }=0$, i.e. $\beta =0$.

Similarly, looking at the $r^{\frac{10 \alpha +5 \beta + \sqrt{100 \alpha ^2 + 84 \alpha \beta + 17 \beta ^2}}{4 \alpha + 2 \beta }}$ term (the term of (2) in Fig. 3 between the third and fourth line):

$$\begin{aligned} \left( \left( \beta + 4 \alpha \right) \sqrt{ 100 \alpha ^2+ 84 \alpha \beta + 17 \beta ^2} + \left( 2 \alpha + \beta \right) \left( 7 \beta + 20 \alpha \right) \right) C_1=0. \end{aligned}$$

(79)

We once more rephrase this as:

$$\begin{aligned} C_1 \left( \left( 2-\frac{\beta }{\chi } \right) \sqrt{ 25 - 8 \frac{\beta }{\chi } } +10- 3 \frac{\beta }{\chi } \right) =0, \end{aligned}$$

which implies that $C_1=0$ or $\frac{\beta }{\chi }=3$.

Thus outside of $\frac{\beta }{\chi } = 2,$ $\frac{25}{8}$ $\frac{21}{8}$, $\frac{28}{9}$ 0, 3, one must have $C_1=C_2=0$, which implies that the metric is Schwarzschild-de Sitter (or AdS). We only have to test these remaining values to conclude. For convenience, and in order to use the same Maple procedure, we will rephrase those in term of $\alpha$ and $\beta$. We need to test the cases: $\beta +4 \alpha =0$, $50 \alpha + 17 \beta =0$, $42 \alpha +13 \beta =0$, $56 \alpha + 19 \beta =0$, $\beta =0$, $3 \alpha + \beta =0$. Actually, this last case corresponds to the conformally invariant one and will not be considered here (see Proposition 4.2 for this configuration).

On the Maple results displayed in Fig. 5, one can see that for $\beta +4 \alpha =0$, $42 \alpha +13 \beta =0$, $56 \alpha + 19 \beta =0$ one must have $C_1=C_2=0$, which is the desired result. In the configuration $50 \alpha + 17 \beta =0$ however, one obtains only $C_1+C_2=0$. Nevertheless, since this corresponds to the case where $f=g$, one concludes that $M(r)= m + \frac{\Lambda }{3}r^3$ (second line of (4) in Fig. 5), which implies that the metric is indeed Schwarzschild-de Sitter (or AdS).

Fig. 5

Cases $\beta +4 \alpha =0$, $50 \alpha + 17 \beta =0$, $42 \alpha +13 \beta =0$, $56 \alpha + 19 \beta =0$

In the final case: $\beta = 0,$ while $C_1=0$, a priori $\Lambda =0$, $C_2 \ne 0$ is an admissible solution, corresponding to the Reissner-Nordström metric. We can check that it is indeed a solution (see Fig. 6) and conclude the proof.

Fig. 6

The Reissner-Nordström metric is $A_{\alpha , 0}$-flat

Of course the above stands when $100 \alpha ^2 + 84 \alpha \beta + 17 \beta ^2\ge 0$. The reasoning when $100 \alpha ^2 + 84 \alpha \beta + 17 \beta ^2<0$ will be very similar. We will thus give a brief overview of the proof in that case: one simply has to replace f and g by

$$\begin{aligned} f&= \frac{6 \alpha +3 \beta + i\sqrt{\left| 100 \alpha ^2 + 84 \alpha \beta + 17 \beta ^2 \right| }}{2\left( 2 \alpha + \beta \right) } \\ g&= \frac{6 \alpha +3 \beta - i\sqrt{\left| 100 \alpha ^2 + 84 \alpha \beta + 17 \beta ^2\right| }}{2\left( 2 \alpha + \beta \right) }. \end{aligned}$$

The algebraic operations will remain the same even with complex exponents, and thus A will be written as a sum of (complex) powers of r. One simply has to replace the $t_i^{\pm }$ by:

$$\begin{aligned} t_1^-&=\frac{18 \alpha +9 \beta - i\sqrt{|100 \alpha ^2 + 84 \alpha \beta + 17 \beta ^2|}}{4 \alpha + 2 \beta }\\ t_2^-&=\frac{10 \alpha +5 \beta - i\sqrt{|100 \alpha ^2 + 84 \alpha \beta + 17 \beta ^2|}}{4 \alpha + 2 \beta }\\ t_3^-&=\frac{6 \alpha +3 \beta - i\sqrt{|100 \alpha ^2 + 84 \alpha \beta + 17 \beta ^2|}}{4 \alpha + 2 \beta }\\ t_4^-&=\frac{6 \alpha +3\beta - i\sqrt{|100 \alpha ^2 + 84 \alpha \beta + 17 \beta ^2|}}{2 \alpha + \beta }\\ t_1^+&=\frac{18 \alpha +9 \beta +i\sqrt{|100 \alpha ^2 + 84 \alpha \beta + 17 \beta ^2|}}{4 \alpha + 2 \beta }\\ t_2^+&=\frac{10 \alpha +5 \beta + i\sqrt{|100 \alpha ^2 + 84 \alpha \beta + 17 \beta ^2|}}{4 \alpha + 2 \beta }\\ t_3^+&=\frac{6 \alpha +3 \beta +i\sqrt{|100 \alpha ^2 + 84 \alpha \beta + 17 \beta ^2|}}{4 \alpha + 2 \beta }\\ t_4^+&=\frac{6 \alpha +3\beta +i\sqrt{|100 \alpha ^2 + 84 \alpha \beta + 17 \beta ^2|}}{2 \alpha + \beta }. \end{aligned}$$

In this case, the $t_i^{\pm }$ cannot interfere and thus the $r^{t_i^{\pm }}$ form a free family. One can then, mutatis mutandis, look at (78) and (79) in the same manner as before, and conclude that $C_1=C_2=0$, and thus that the metric is Schwarzschild-de Sitter (or AdS).