20
$\begingroup$

This is my first question on this site. I hope it is not inappropriate on MO.

Myers and Steenrod proved 1939 that the isometry group of a Riemannian manifold is a lie group. I add a picture where Kobayashi describes the base idea of this proof.

enter image description here

Now I read in the book "Recent trends about lorentzian geometry" the following: enter image description here (The reference [6] is Myers and Steenrod).

Sadly the author doesn't give more details. Can anyone tell my why this particular proof of M&S does not work for Pseudo-Riemannian manifolds?

EDIT: Maybe this will clear things up a bit.

(1) My questions is NOT about why theorem 1 (on page 278) fails for Lorentzian manifolds. (Theorem 1 is: If $(M,g)$ is a compact Riemannian manifold, then $Iso(M,g)$ is compact.)

(2) I know that the theorems provided by Kobayashi in his book "Transformation groups in differential geometry" can be applied to Lorentzian manifolds as well (e.g. Thm 5.1). So I wonder why this specific proof of M&S only works for Riemannian manifolds.

Improve this question
$\endgroup$
8
  • 1
    $\begingroup$ mathoverflow.net/questions/141417/… $\endgroup$
    – Carlo Beenakker
    Commented Oct 22, 2016 at 18:34
  • 1
    $\begingroup$ @CarloBeenakker: Thanks for the link. I read that question before I wrote mine and I also read the proof of Kobayashi where he embeds the isometry group in the bundle of orthonormal frames as a closed submanifold. I know that this works for Pseudo-Riemannian manifolds too. So I wondered why the original proof from Myers and Steenrod doesn't. $\endgroup$
    – JS.
    Commented Oct 22, 2016 at 18:42
  • 2
    $\begingroup$ I just took a look in this book you mention, "Recent trends about lorentzian geometry" ", I presume you are referring to the article "On the Isometry Group of Lorentz Manifolds". There it says that the theorem that fails in the Lorentzian case is "If (M,g) is a compact Riemannian manifold, then Iso(M,g) is compact." --- which seems an altogether different kettle of fish. If this is not the paper you had in mind, perhaps a more specific pointer will help. $\endgroup$
    – Carlo Beenakker
    Commented Oct 22, 2016 at 19:31
  • 2
    $\begingroup$ Yes this is the paper I had in mind, but I refer to the introduction on page 278, where the author says "We point out that Iso(M,g) has a Lie group structure when considered with the compact-open topology. For Riemannian metrics, this has been established (long ago) in [6]. However, the techniques employed there do not generalize to semi-Riemannian metrics." (The reference [6] is Myers and Steenrod). $\endgroup$
    – JS.
    Commented Oct 22, 2016 at 21:00
  • $\begingroup$ What you refer to is theorem 1 on the same page. This is a direct conclusion of the embedding into the bundle of orthonormal frames as closed submanifold. Is $(M,g)$ Riemannian and compact then the orthonormal frame bundle is compact since it has the orthogonal group as fibers which is compact. However this is not true in general for Pseudo-Riemannian manifolds. But my question does not refer to this theorem. $\endgroup$
    – JS.
    Commented Oct 22, 2016 at 21:09

1 Answer 1

Reset to default
6
$\begingroup$

I think the main reason is basically Myers and Steenrod use properties of Riemannian manifolds related to their other theorem of differential geometry by the same name in the same 1939 paper (but not involving Lie groups) for distance function and isometries. Since this other theorem doesn't generalize to Lorentzian manifolds for obvious reasons, their proof of their theorem for Lie groups using those arguments is not generalizable to Lorentzian manifolds. The usual modern proofs(by Sternberg and Kobayashi for instance) avoid arguments related to distance functions by using local flows and orthonormal basis bundles.

Improve this answer
$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .