Myers-Steenrod states that the isometry group of a Riemannian manifold is a Lie group. Is that also true for pseudo Riemannian manifolds? I didn't find anything related to that.
Cheers
Isometry Group – Pseudo Riemannian Manifold as Lie Group?
dg.differential-geometryisometrieslie-groupsriemannian-geometry
Related Question
- [Math] Is every topological (resp. Lie-) group the isometrygroup of a metric space (resp. Riemannian manifold)
- [Math] Why does the proof of Myers and Steenrod fail in the Lorentzian case
- Pseudo-Riemannian Manifold Without Nonzero Null Vectors – Terminology
- Differential Geometry – Complexified Isometry Group Acting on Tangent Bundle of Compact Riemannian Manifold
Best Answer
Yes. Check out Kobayashi, Transformation Groups in Differential Geometry, theorem 4.1 page 16, and example 2.5 page 8. The automorphisms of a pseudo-Riemannian manifold form a Lie group, as do the automorphisms of a conformal pseudo-Riemannian manifold (in dimension 3 or more), and the automorphisms of a projective connection.
The basic idea of the proof is to show that the bundle of orthonormal frames has a canonical basis of global 1-forms (expressed in terms of the Levi-Civita connection of the pseudo-Riemannian metric). Then you show that no diffeomorphism of a manifold can fix a point and also fix a basis of global 1-forms. You use this to show that the automorphism group actually immerses into the bundle of orthonormal frames, by taking frame $\phi$ and mapping each automorphism $g$ to $g\phi$. The automorphism group orbit equals the orthonormal bundle precisely if the manifold is a homogeneous pseudo-Riemannian space form.