Isometry group of pseudo Riemannian manifold always a Lie group? (Myers-Steenrod) 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
 A: 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 $\varphi$ and mapping each automorphism $g$ to $g\varphi$. The automorphism group orbit maps onto the orthonormal frame bundle precisely if the manifold is a homogeneous pseudo-Riemannian space form.
Careful: Kobayashi proves that, in a suitable smooth structure, the automorphism group injectively immerses into the bundle of orthonormal frames with closed image. He claims without proof that it is an embedding. This is true, but requires properness, which one can obtain from Myers-Steenrod by putting a suitable invariant Riemannian metric on the frame bundle. I am writing up some notes on Cartan geometries with a more detailed proof. This problem has been noted before, in particular here: Palais's and Kobayashi's theorems on automorphism groups of geometric structures
