Can a spacetime solution in GR have no Killing vector fields?

Sometimes Killing vector fields in a given spacetime are described as giving information about a symmetry of that particular spacetime solution.

If I look at the requirement of a Killing vector field (is this a sufficient requirement, or is there more?):

$$\nabla_\mu X_\nu + \nabla_\nu X_\mu = 0$$

This appears purely local, and because General Relativity is a diffeomorphism invariant theory, this makes me wonder if spacetime solutions will necessarily have an infinite number of "local" Killing vector fields (where "local" means the vector field is only non-zero within a finite region of spacetime, and zero outside of this).

Question: what kinds of symmetries (local, global, continuous, discrete, etc.) can be associated with a Killing vector field?

is it possible for a spacetime solution in GR to have none of these symmetries, and so admit zero Killing vector fields?

• Answers in a related question: physics.stackexchange.com/q/98119 explain that there are at least some global symmetries (discrete ones) which are not describable with a Killing vector field. – Student4life Dec 21 '14 at 2:31
• The equation for the Killing field should use covariant derivatives. The notation you used in the question implies ordinary coordinate derivatives. – Philip Gibbs - inactive Dec 21 '14 at 8:08
• @PhilipGibbs I thought $\nabla_\mu$ for covariant derivatives and $\partial_\mu$ for ordinary coordinate derivatives, was standard notation. I'm sorry if the notation wasn't clear, feel free to edit the question if you feel it needs updating. – Student4life Dec 25 '14 at 6:21
• @Studemt4life I've not seen the notation used that way before but fair enough if that was what you intended. I have seen $D_\mu$ used in that way – Philip Gibbs - inactive Dec 25 '14 at 9:10
• @PhilipGibbs I've only seen $D_\mu$ used for gauge covariant derivatives, like in QED or the standard model. I didn't realize these had such different meanings to people. Is the "," or ";" notation for derivatives more "universal"? Seems to be the only notation we've both seen used the same way. Maybe I'll stick to that in the future. – Student4life Dec 26 '14 at 6:32

The existance of Killing fields even just for a small region is a special property of the metric. For general metrics you cannot expect to find Killing vectors. Notice that the Killing equation which should be written using covariant derivatives as $X_{\mu;\nu} + X_{\nu;\mu} = 0$ is 10 independent partial differential equations for only 4 field components so solutions should not be expected in general. In other words the typical case is that a metric has no symmetries.