A new non-inheriting homogeneous solution of the Einstein-Maxwell equations with cosmological term

Abstract

We find a new homogeneous solution to the Einstein-Maxwell equations with a cosmological term. The spacetime manifold is $\mathbf{R}\times \mathbf{S}^3$. The spacetime metric admits a simply transitive isometry group $G = \mathbf{R} \times \mathbf{SU(2)}$ and is Petrov type I. The spacetime is geodesically complete and globally hyperbolic. The electromagnetic field is non-null and non-inheriting: it is only invariant with respect to the SU(2) subgroup and is time-dependent in a stationary reference frame.

A Petrov type

In this appendix we give the details of the determination of the Petrov type of the solution.

According to the results, e.g., in references [1, 16], in order to prove that the spacetime investigated here has Petrov type I it is sufficient to establish that

$$\begin{aligned} I^3 - 27 J^2 \ne 0, \end{aligned}$$

(A.1)

where the scalar invariants I and J are given in terms of the Newman-Penrose Weyl scalars by

$$\begin{aligned} I&= \psi _0\psi _4 - 4\psi _1\psi _3 + 3\psi _2^2 \end{aligned}$$

(A.2)

$$\begin{aligned} J&= \psi _0\left( \psi _2\psi _4 - \psi _3^2\right) +\psi _2\left( 2\psi _1\psi _3 - \psi _2^2\right) -\psi _4\psi _1^2. \end{aligned}$$

(A.3)

Using the null tetrad constructed from the vector fields in (4.3) according to

$$\begin{aligned} k = \frac{1}{\sqrt{2}}(e_0 + e_1),\quad l =\frac{1}{\sqrt{2}}(e_0 - e_1),\quad m =\frac{1}{\sqrt{2}}(e_2 + i e_3), \quad \overline{m} = \frac{1}{\sqrt{2}}(e_2 - i e_3), \end{aligned}$$

(A.4)

the Weyl scalars are given by

$$\begin{aligned} \psi _0&= \frac{(B - C)}{4ABC\beta ^2} \left[ B+C-A-\sqrt{A}\right] , \end{aligned}$$

(A.5)

$$\begin{aligned} \psi _1&=0, \end{aligned}$$

(A.6)

$$\begin{aligned} \psi _2&=\frac{1}{12AB C \beta ^2} \left[ 2A^2 - (B+C+2)A- (B-C)^2\right] , \end{aligned}$$

(A.7)

$$\begin{aligned} \psi _3&=0, \end{aligned}$$

(A.8)

$$\begin{aligned} \psi _4&= \frac{(B - C)}{4ABC\beta ^2} \left[ B+C-A+\sqrt{A}\right] . \end{aligned}$$

(A.9)

Using (3.39) we obtain

$$\begin{aligned}&I^3 - 27 J^2 \nonumber \\&\quad =- 4\frac{(187804027) (3^2) (3557) 2^{2/3} +(369518691691) (41) 2^{-2/3} + (1051)(9071602009)}{5^{24}\beta ^{12}}\nonumber \\&\quad \ne 0. \end{aligned}$$

(A.10)