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?

  • $\begingroup$ 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. $\endgroup$ Dec 21, 2014 at 2:31
  • $\begingroup$ The equation for the Killing field should use covariant derivatives. The notation you used in the question implies ordinary coordinate derivatives. $\endgroup$ Dec 21, 2014 at 8:08
  • $\begingroup$ @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. $\endgroup$ Dec 25, 2014 at 6:21
  • $\begingroup$ @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 $\endgroup$ Dec 25, 2014 at 9:10
  • $\begingroup$ @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. $\endgroup$ Dec 26, 2014 at 6:32

3 Answers 3


Killing vectors are infinitesimal isometries - they are determined by taking the Lie derivative of the metric tensor with respect to a vector field, e.g., $\mathcal {L}_Xg=0$ where $X$ is a local vector field.

For Minkowski space, $$ds^2=-dt^2+dx^2+dy^2+dz^2,$$ there are $(4)$ infinitesimal vector generators for translations: $$\partial_t,\partial_x,\partial_y,\partial_z,$$ there are $(3)$ infinitesimal vector generators for Lorentz rotations: $$-y\partial_x + x \partial_y,z\partial_y+y\partial_z,z\partial_x-x\partial_z,$$ and there are $(3)$ infinitesimal vector generators for Lorentz boosts: $$x\partial_t + ct \partial_x, y\partial_t + ct \partial_y, z\partial_t + ct \partial_z.$$

The maximum number of Killing vectors for Minkowski space is $10$ - which is the maximum number of Killing vectors for any semi-Riemannian manifold of dimension $4$ with Lie orthogonal group $O(4)$ of dimension of $6.$

If you need to restrict the manifold, then it may be possible to find conformal Killing vectors where $X$ defines a conformal tranformation, e.g., $\mathcal {L}_Xg=\lambda g$ where $\lambda$ is a function which preserves $g$ up to scale and the conformal structure.

And yes, it's possible to have a metric which doesn't admit any Killing vectors under the flow of the vector field $X$.


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.

You are right that the field equations for the metric are diffeomorphism invariant but solutions of differential equations with a given symmetry do not usually preserve any part of that symmetry unless the initial conditions also do.

The question about which kind of symmetries are possible is equivalent to the problem of classifying symmetric spaces for which I refer you to Wikipedia https://en.wikipedia.org/wiki/Symmetric_space#Classification_results

  • $\begingroup$ What about local lorentz symmetry? It is a continuous symmetry, so isn't ruled out by the linked answer in the comments. A coordinate system can always be chosen so that the metric is nice and +/- 1 diagonal along any timelike path. The metric would be invariant to a lorentz boost coordinate change. Since the equations for a killing field is coordinate system independent, we are talking about the symmetry of the geometry, which therefore must include these lorentz symmetries along any timelike path. Why do these symmetries not lead to a killing field? What symmetries "count"? $\endgroup$ Dec 25, 2014 at 6:30
  • $\begingroup$ @Student4life Local lorentz symmetry is a symmetry in the dynamical field equations, but a Killing field is giving you a symmetry in the metric treated as a fixed backgound. The Killing field equation needs to apply everywhere. It does not give any kind of symmetry if it holds only along a worldline. $\endgroup$ Dec 25, 2014 at 9:14
  • $\begingroup$ All GR solutions have local lorentz symmetry (due to it being a lorentzian manifold). Can't the killing field just be zero outside of some local patch (as being zero would trivially satisfy the derivative requirement)? You seem to be saying no, but without explaining why. These are exactly the issues I was hoping an answer would address: What kinds of symmetries (local, global, continuous, discrete, etc.) can be associated with a Killing vector field? $\endgroup$ Dec 26, 2014 at 6:37
  • $\begingroup$ Even though the defining equation for a killing vector field is a local equation, you seem to be saying that a global symmetry is required. Why? $\endgroup$ Dec 26, 2014 at 6:43
  • $\begingroup$ @Student4life I think your definition of local symmetry may be different from mine so you will need to give your definition to clear up any confusion. My understanding of local symmetry in GR is diffeomorphism invarinace. A background field could only have diffeomorphism invariance if every vector field was a Killing field and that is not possible unless the metric is zero everywhere. $\endgroup$ Dec 26, 2014 at 9:02

The symmetries Killing fields are associated with are continuous isometries specifically. Assuming the isometry group of a (pseudo-)Riemannian manifold is a Lie group of positive dimension, then the Killing fields are closely related to its associated Lie algebra, in fact, they form a Lie algebra that is isomorphic to it. Generically, Riemannian manifolds only have the trivial isometry as a symmetry, which being a discrete group, means that a generic Riemannian manifold only has the 0 vector field as a Killing field, which I suppose technically is a Killing field, but not a very interesting one.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.