9
$\begingroup$

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

$\endgroup$
12
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ 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 '16 at 12:49
  • $\begingroup$ Nevermind, reading further along theorem 5.1 page 22 gives the general case. $\endgroup$ – iolo Jul 8 '16 at 13:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.