# Is the Kerr metric more symmetric than a normal type D spacetime?

The Kerr spacetime is of Petrov type D (see here for the Petrov classification of spacetimes). In the Newman-Penrose formalism, from the Goldberg-Sachs theorem we can conclude that there is a choice of null tetrad such that the following Newman-Penrose scalars are zero

$$\begin{equation} \kappa=\sigma=\nu=\lambda=0, \end{equation}$$

in addition to the Weyl scalars $$\Psi_0=\Psi_1=\Psi_3=\Psi_4=0$$ from the spacetime being type D.

For the Kerr spacetime, one can choose the Kinnersly tetrad such that the Newman-Penrose scalar $$\epsilon=0$$ in addition to the above quantities. Does this indicate that the Kerr spacetime is in some sense "more symmetric" than a "typical" type D spacetime? Or in general can one set one more Newman-Penrose scalar to zero in a type D spacetime beyond the scalars that can be zero due to the Goldberg-Sachs theorem?

EDIT: in addition to the answer given below, reading through Kinnersley's original paper I realize that in fact he sets $$\epsilon=0$$ for a general type D metric.

• Well, essentially all Petrov D space-times are in the Plebanski-Demianski class, and Kerr is the only asymptotically flat and vacuum one...
– Void
Nov 15, 2019 at 0:10
• @Void why would being asymptotically (A)dS make it any less symmetric? Nov 15, 2019 at 8:10
• @mmeent It is more the "vacuum" I am talking about, of course once you allow for electro+$\Lambda$-vacuum (with appropriate regular asymptotics) then the class gets enlarged to Kerr-Newman-(A)dS. The asymptotic flatness (regular asymptotics) refers more to eliminating NUT and acceleration.
– Void
Nov 15, 2019 at 12:41

Choosing a null tetrad that satisfies $$\Psi_0=\Psi_1=\Psi_3=\Psi_4=0$$ in a Petrov type D spacetime requires only 4 of the 6 degrees of freedom in choosing a null tetrad. One remaining degree of freedom is the ability to use a boost to rotate the two principle null vectors ($$l^\mu$$ and $$n^\mu$$ in usual notation) into each other. The other is the freedom to rotate the other two legs of the tetrad ($$m^\mu$$ and $$\bar{m}^\mu$$) leaving $$l^\mu$$ and $$n^\mu$$ invariant. These freedoms can always be used to set $$\epsilon=0$$, for any Petrov type D spacetime. So this particular property does not set Kerr apart.