# Highest symmetric non-maximally symmetric spacetime

What is the highest number of symmetries (Killing vectors) that a (4-dimensional) spacetime can have without being maximally symmetric? From what I can see, it seems to be 7 (which includes the Einstein universe and some pp-wave spacetimes), but the theorems used (in chapter 11 and 12 of Stephani's "Exact Solutions of the Einstein's Field Equations") only apply to spacetimes stemming from a variety of stress energy tensors (Vacuum, lambda vacuum, EM fields, perfect fluids, pure radiation).

Can that result be generalized to all spacetimes, regardless of their sources?

• Please put the theorems here for those who do not have access to the text. – Ryan Unger Aug 17 '15 at 0:12

The submaximal dimension of the group of isometries of a Pseudo-Riemanniann manifold of dimension $n$ with $n\ge4$ and $n\neq5$ is $$\frac{1}{2}n(n-1)+ 2 .$$

However, a result proved here(Theorem 3.2) shows that a spacetimes with that amount of isometries in dimension $4$ must be Minkoswki spacetime.

Hence, the maximal number of Killing vectors you can have (without the trivial Minkoswki one) is $\bf{7}$.

Part of the main results of the paper are:

Let $l_{0}(n) > l_{1}(n) >...$ be the possible dimensions of all groups of isometries of Lorentz manifolds, listed in their decreasing order. Such dimensions are called Lorentz degrees of symmetries. We say that an n-dimensional connected Lorentz manifold $M$, belongs to the j-stratum and write $M\in L_{j}(n)$; if there is a Lie group $K$; $dim K = l_{j}(n)$, that acts effectively on $M$ by isometries.

We show that the only Lorentz manifolds in $L_{0}(n)$ are the Minkowski space and the Wolf spaces and for $n\ge4$; $n \ne 5$ the only manifold in $L_{1}(n)$ is Minkowski space.

Also you can look the following paper for additional information.