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.



1 Answer 1


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.

  • $\begingroup$ I am struggling to understand the reference. Isn't theorem 4.1 proved only for compact manifolds. (I am a not good at dg, so I may be wrong) $\endgroup$
    – iolo
    Jul 8, 2016 at 12:49
  • $\begingroup$ Nevermind, reading further along theorem 5.1 page 22 gives the general case. $\endgroup$
    – iolo
    Jul 8, 2016 at 13:17

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