On the Goldberg–Sachs theorem in six dimensions

Abstract

We study the “shear-free” part of the Goldberg–Sachs theorem for type II and D six-dimensional Einstein spacetimes. A spacetime is of the type II or D if it admits a double Weyl aligned null direction (WAND). The Goldberg–Sachs theorem then restricts geometric properties of double WANDs. This restriction can be expressed in terms of constraints on the form of the “optical matrix” defining the expansion, rotation, and shear of the double WAND. We show that a rank-four and rank-three optical matrix is orthogonally similar to one of three canonical forms, reducing the number of free parameters of the optical matrix from ten to five.

A Solution to \(\mu = 0\)

We define $\mu$ as in (79a)

$$\begin{aligned} \mu = \left( \alpha \beta \gamma + \alpha q^{2} + \beta s^{2} + \gamma u^{2}\right) ^{2} - 4 \alpha \gamma (\beta s + qu)^{2}, \end{aligned}$$

(89)

where

$$\begin{aligned} \alpha&= t - p, \end{aligned}$$

(90a)

$$\begin{aligned} \beta&= v - p,\end{aligned}$$

(90b)

$$\begin{aligned} \gamma&= v - t. \end{aligned}$$

(90c)

Let us discuss solutions of the equation

$$\begin{aligned} \mu = 0. \end{aligned}$$

(91)

Note that

$$\begin{aligned} \alpha&\ge 0, \end{aligned}$$

(92a)

$$\begin{aligned} \gamma&\ge 0,\end{aligned}$$

(92b)

$$\begin{aligned} \beta&= \alpha + \gamma . \end{aligned}$$

(92c)

Proposition 4

For each solution of the equation $\mu = 0$, at least one of the following conditions is fulfilled

$$\begin{aligned} \alpha = \gamma = 0, \end{aligned}$$

(93a)

$$\begin{aligned} q = s = \gamma = 0, \end{aligned}$$

(93b)

$$\begin{aligned} s = u = \alpha = 0, \end{aligned}$$

(93c)

$$\begin{aligned} q = u = 0,\quad s^{2} = \alpha \gamma , \end{aligned}$$

(93d)

$$\begin{aligned} \alpha q^{2} = \gamma u^{2} = qsu \ne 0. \end{aligned}$$

(93e)

Proof

Let us suppose that none of (93a)–(93d) holds. We will show that then (93e) must hold.

It is straightforward to see that $\mu = 0$ implies

$$\begin{aligned} q = \gamma = 0 \quad&\Rightarrow \quad s = 0, \end{aligned}$$

(94a)

$$\begin{aligned} u = \alpha = 0 \quad&\Rightarrow \quad s = 0, \end{aligned}$$

(94b)

$$\begin{aligned} q = u = 0 \quad&\Rightarrow \quad s^2 = \alpha \gamma . \end{aligned}$$

(94c)

Therefore, under the given assumptions,

$$\begin{aligned} \alpha q^{2} + \gamma u^{2} > 0. \end{aligned}$$

(95)

Treating $\mu$ as a quadratic polynomial in $\alpha$ and $\gamma$, we see that each time, the discriminant is non-positive, so we complete the square

$$\begin{aligned} (\beta \gamma + q^{2})^{2} \mu&= h_{1}^{2} + 4 \beta \gamma k_{1}^{2}, \end{aligned}$$

(96a)

$$\begin{aligned} (\alpha \beta + u^{2})^{2} \mu&= h_{2}^{2} + 4 \alpha \beta k_{2}^{2}, \end{aligned}$$

(96b)

where

$$\begin{aligned} h_{1}&= \alpha (\beta \gamma + q^{2})^{2} + (\gamma u^{2} - \beta s^{2})(\beta \gamma - q^{2}) - 4 \beta \gamma qsu,&k_{1}&= (\beta s + qu)(\gamma u - qs), \end{aligned}$$

(97a)

$$\begin{aligned} h_{2}&= \gamma (\alpha \beta + u^{2})^{2} + (\alpha q^{2} - \beta s^{2})(\alpha \beta - u^{2}) - 4 \alpha \beta qsu,&k_{2}&= (\beta s + qu)(\alpha q - su). \end{aligned}$$

(97b)

As both $\beta \gamma$ and $\alpha \beta$ are strictly positive, four equalities follow from $\mu = 0$

$$\begin{aligned} h_{1} = h_{2} = k_{1} = k_{2} = 0. \end{aligned}$$

(98)

Choosing an appropriate combination of $h_{1}$ and $h_{2}$ gives

$$\begin{aligned} 0 = \alpha h_{1} - \gamma h_{2} = (\alpha \beta \gamma + \alpha q^{2} + \beta s^{2} + \gamma u^{2})(\alpha q^{2} - \gamma u^{2}). \end{aligned}$$

(99)

The first factor is strictly positive due to (95) and thus

$$\begin{aligned} \alpha q^{2} = \gamma u^{2}. \end{aligned}$$

(100)

Hence, considering (95) again,

$$\begin{aligned} \alpha \gamma qu \ne 0. \end{aligned}$$

(101)

Considering $k_{1} = 0$, $k_{2} = 0$, one of the following two conditions must hold

$$\begin{aligned} \beta s + qu&= 0\quad (\text {hence }qsu < 0), \end{aligned}$$

(102a)

$$\begin{aligned} (\gamma u - qs)^{2} + (\alpha q - su)^{2}&= 0\quad (\text {hence }qsu > 0). \end{aligned}$$

(102b)

However, it can be seen from (89) and (95) that it is impossible to achieve (102a). $\square$

Corollary 1

The equation

$$\begin{aligned} \mu = 0,\quad \alpha ^{2} + \gamma ^{2} \ne 0 \end{aligned}$$

(103)

has the following set of roots, parametrized by real x, y, z,

$$\begin{aligned} \alpha&= y^{2},\end{aligned}$$

(104a)

$$\begin{aligned} \gamma&= x^{2},\end{aligned}$$

(104b)

$$\begin{aligned} q&= xz,\end{aligned}$$

(104c)

$$\begin{aligned} s&= xy,\end{aligned}$$

(104d)

$$\begin{aligned} u&= yz. \end{aligned}$$

(104e)

Proof

To see that (104) solves $\mu = 0$, one can substitute into (89)

$$\begin{aligned} \left( \alpha \beta \gamma + \alpha q^{2} + \beta s^{2} + \gamma u^{2}\right) ^{2} = 4 \alpha \gamma (\beta s + qu)^{2} = 4 x^{4} y^{4} \left( x^{2} + y^{2} + z^{2}\right) ^{2}. \nonumber \\ \end{aligned}$$

(105)

It is obvious that cases (93b)–(93d) can be parametrized according to (104). In order to complete the proof, it suffices to show that this parametrization can be also applied to the case (93e).

Let us assume (93e)

$$\begin{aligned} \alpha q^{2} = \gamma u^{2} = qsu \ne 0. \end{aligned}$$

(106)

Hence

$$\begin{aligned} \begin{pmatrix} -s &{}\quad \gamma \\ \alpha &{}\quad -s \end{pmatrix} \begin{pmatrix} q\\ u \end{pmatrix} = \begin{pmatrix} 0\\ 0 \end{pmatrix}. \end{aligned}$$

(107)

The determinant necessarily vanishes

$$\begin{aligned} s^{2} - \alpha \gamma = 0. \end{aligned}$$

(108)

Thus one can introduce x and y such that

$$\begin{aligned} \alpha&= y^{2},\end{aligned}$$

(109a)

$$\begin{aligned} \gamma&= x^{2},\end{aligned}$$

(109b)

$$\begin{aligned} s&= xy. \end{aligned}$$

(109c)

By substituting this parametrization into (107), we get

$$\begin{aligned} \begin{pmatrix} x\\ -y \end{pmatrix} \begin{pmatrix} -y&x \end{pmatrix} \begin{pmatrix} q\\ u \end{pmatrix} = \begin{pmatrix} 0\\ 0 \end{pmatrix}, \end{aligned}$$

(110)

which implies

$$\begin{aligned} qy = ux. \end{aligned}$$

(111)

Finally, one can introduce z such that

$$\begin{aligned} q&= xz,\end{aligned}$$

(112a)

$$\begin{aligned} u&= yz. \end{aligned}$$

(112b)

$\square$