# Must every isometry have an associated Killing vector?

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I understand that the flows of Killing vector fields are isometries, and that one-parameter groups of isometries have an associated Killing vector which generates them, but are your Killing vectors guaranteed to give you all possible isometries of the manifold? I guess what I'm trying to ask is, do isometries necessarily come in families, or can you have "isolated" isometries which can't have an associated Killing vector field?

This post imported from StackExchange Physics at 2014-04-11 15:21 (UCT), posted by SE-user JohnnyMo1

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The group of isometries of a given connected smooth (semi) Riemannian manifold is always a Lie group. However, a Lie group can include subgroups of discrete isometries that, barring the identity, cannot be represented by continuous isometries and thus they have no Killing vectors associated with them. (Actually, only some elements of the connected component including the identity can be associated to Killing fields.)

For instance, referring to $\mathbb R^3$ equipped with the standard metric, the Lie group of isometries is the semidirect product of space translations $\mathbb R^3$ and rotations $O(3)$ around a fixed point. The second mentioned subgroup of isometries, $O(3)$, admits a discrete subgroup: $\{I, -I\}$. The spatial inversion $-I$ cannot be associated with any Killing field. Similarly all the symmetries in $-I(SO(3))$ cannot be associated with Killing fields.

This post imported from StackExchange Physics at 2014-04-11 15:21 (UCT), posted by SE-user V. Moretti
answered Feb 6, 2014 by (2,075 points)
Interesting. I think I can intuit that those discrete isometries can't be part of the connected component of the identity because then they'd clearly be part of some family, but can other connected components of the isometry group be associated with Killing vectors and is there some way to tell which ones? Or can you not have "big" connected components outside the identity, just little discrete isometry points?

This post imported from StackExchange Physics at 2014-04-11 15:21 (UCT), posted by SE-user JohnnyMo1
Actually I never considered these issues. If the Lie group of isometries is compact, then every element in the connected component of the identity belongs to a one-parameter subgroup and thus, acting on the manifold, gives rise to a Killing field. If the group is not compact, it is possible that there are points in the said connected component that do not stay one any one parameter subgroup (though they are connected to the identity via a piecewise smooth curve). In that case I am not sure that these isometries are associated to Killing fields.

This post imported from StackExchange Physics at 2014-04-11 15:21 (UCT), posted by SE-user V. Moretti
Consider Minkowski spacetime. Lorentz group is part of the isometries group and it is not compact. A generic transformation of the form $\Lambda R$, where $\Lambda$ is a boost and $R$ a $3$-rotation, does not belong to any one parameter subgroup. It is not associated to a Killing vector, whereas, separately, $\Lambda$ and $R$ are.

This post imported from StackExchange Physics at 2014-04-11 15:21 (UCT), posted by SE-user V. Moretti
Elements of other connected components different from that including the identity cannot be associated with Killing vectors. This is because to define a Killing vector at $p$ you have to move $p$ under the action of smooth curve of transformations of the Lie group of isometries that passes through the identity (when $p$ is fixed). This curve can therefore only passes through points belonging the connected component of the identity by construction.

This post imported from StackExchange Physics at 2014-04-11 15:21 (UCT), posted by SE-user V. Moretti
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Yes, you can have such "isolated" isometries.

Consider the real line $\mathbb R$ and the inversion mapping $x\to -x$. This isometry does not arise from a killing vector because it's not "continuously connected to the identity."

This post imported from StackExchange Physics at 2014-04-11 15:21 (UCT), posted by SE-user joshphysics
answered Feb 6, 2014 by (835 points)

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