Do there exist a nontrivial Killing field on each riemannian manifold?
A Killing field is a vector field whose flow acts on the manifold by isometry.
Asked
Active
Viewed 1,095 times
1 Answers
7
The isometry group of a Riemannian manifold is a finite-dimensional Lie group. A Killing field defines a 1-parameter subgroup $\Bbb R \to \text{Isom}(M)$ by flowing; and conversely, taking the derivative of a 1-parameter subgroup $\Bbb R \to \text{Isom}(M)$ at $e$ gives a Killing field.
By exponentiating, one sees that 1-parameter subgroups of a Lie group are in bijection with elements of the Lie algebra $\mathfrak g$. So a Riemannian manifold has a nontrivial Killing field if and only if $\text{Isom}(M)$ is positive-dimensional. For the vast majority of Riemannian manifolds, this is not true - I don't know how to make this rigorous, but a 'generic' Riemannian manifold should have no nontrivial isometries whatsoever.
But if that's unsatisfying, you can have otherwise very nice Riemannian manifolds - say, surfaces of genus $g>1$ with constant negative curvature - whose isometry group is finite. In full generality any compact, $n>1$-dimensional hyperbolic manifold has finite isometry group, so has no nontrivial Killing field.