Why does the proof of Myers and Steenrod fail in the Lorentzian case?

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.

Now I read in the book "Recent trends about lorentzian geometry" the following: (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.

• mathoverflow.net/questions/141417/… – Carlo Beenakker Oct 22 '16 at 18:34
• @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. – JS. Oct 22 '16 at 18:42
• 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. – Carlo Beenakker Oct 22 '16 at 19:31
• 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). – JS. Oct 22 '16 at 21:00
• 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. – JS. Oct 22 '16 at 21:09