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?
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$\begingroup$ Your wording "isolated" isometries is rather insightful; as in V.Moretti's and Joshphysics's answers, the non-Killing-associated isometries belong to a discrete subgroup. You show considerable mathematical intuition! $\endgroup$– Selene RoutleyJun 16, 2014 at 2:23
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2 Answers
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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.
answered Feb 6, 2014 at 20:28
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$\begingroup$ 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? $\endgroup$ Feb 6, 2014 at 20:39
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2$\begingroup$ 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. $\endgroup$ Feb 6, 2014 at 21:14
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1$\begingroup$ 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. $\endgroup$ Feb 6, 2014 at 21:20
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4$\begingroup$ 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. $\endgroup$ Feb 6, 2014 at 21:34
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1$\begingroup$ Hi Rod. For instance see O'Neill's book on semi Riemannian geometry, one of the last chapters...if the manifold is finite dim the Lie group is finite dim as well. $\endgroup$ Jun 16, 2014 at 6:39
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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."
answered Feb 6, 2014 at 20:30