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.
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Sign up to join this communityMyers-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
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.