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

Question

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?

Answer 1

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 $X$.

Answer 2

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

Answer 3

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.