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Are thereDoes anyone know a good reference for general results aboutabout closed Semi-Riemannian manifolds which have a non-compact isometry group?

Background: By a theorem of Myers and Steenrod the isometry group of a Riemannian manifold is a lie group. The same is true for Semi-Riemannian manifolds where the idea of a possible proof (which also works for Riemannian manifolds) is to embed the isometry group into the bundle of orthonormal frames as a closed submanifold. In the case of Riemannian manifolds this immediately yields that the isometry group of a compact Riemannian manifold has to be compact. But this is in general false for Semi-Riemannian manifolds since the bundle of orthonormal frames is not necessarily compact in this case even if the manifold is. So I ask myself if there are general statements about the phenomenon of a closed Semi-Riemannian manifold with non-compact isometry group.

Edit: My goal is to understand a bit better what the intuition behind compactness / non-compactness of the isometry of a closed Semi-Riemannian manifold is. For example are there topological restrictions on the underlying space.? An example of thisthe kind would beof statements I'm looking for is the following theorem for(for Lorentzian manifolds):

Theorem: Let $(M,g)$ be a closed, simply connected Lorentz manifold and assume $M$ and $g$ are analytic. Then the isometry group of $M$ is compact.

Are there general results about closed Semi-Riemannian manifolds which have a non-compact isometry group?

Background: By a theorem of Myers and Steenrod the isometry group of a Riemannian manifold is a lie group. The same is true for Semi-Riemannian manifolds where the idea of a possible proof (which also works for Riemannian manifolds) is to embed the isometry group into the bundle of orthonormal frames as a closed submanifold. In the case of Riemannian manifolds this immediately yields that the isometry group of a compact Riemannian manifold has to be compact. But this is in general false for Semi-Riemannian manifolds since the bundle of orthonormal frames is not necessarily compact in this case even if the manifold is. So I ask myself if there are general statements about the phenomenon of a closed Semi-Riemannian manifold with non-compact isometry group.

Edit: My goal is to understand a bit better what the intuition behind compactness / non-compactness of the isometry of a closed Semi-Riemannian manifold is. For example topological restrictions on the underlying space. An example of this kind would be the following theorem for Lorentzian manifolds:

Theorem: Let $(M,g)$ be a closed, simply connected Lorentz manifold and assume $M$ and $g$ are analytic. Then the isometry group of $M$ is compact.

Does anyone know a good reference for general results about closed Semi-Riemannian manifolds which have a non-compact isometry group?

Edit: My goal is to understand a bit better what the intuition behind compactness / non-compactness of the isometry of a closed Semi-Riemannian manifold is. For example are there topological restrictions on the underlying space? An example of the kind of statements I'm looking for is the following theorem (for Lorentzian manifolds):

Theorem: Let $(M,g)$ be a closed, simply connected Lorentz manifold and assume $M$ and $g$ are analytic. Then the isometry group of $M$ is compact.

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JS.
  • 893
  • 6
  • 15

Are there general results about closed Semi-Riemannian manifolds which have a non-compact isometry group?

Background: By a theorem of Myers and Steenrod the isometry group of a Riemannian manifold is a lie group. The same is true for Semi-Riemannian manifolds where the idea of a possible proof (which also works for Riemannian manifolds) is to embed the isometry group into the bundle of orthonormal frames as a closed submanifold. In the case of Riemannian manifolds this immediately yields that the isometry group of a compact Riemannian manifold has to be compact. But this is in general false for Semi-Riemannian manifolds since the bundle of orthonormal frames is not necessarily compact in this case even if the manifold is. So I ask myself if there are general statements about the phenomenon of a closed Semi-Riemannian manifold with non-compact isometry group. One statement that I know

Edit: My goal is to understand a bit better what the intuition behind compactness / non-compactness of the isometry of a closed Semi-Riemannian manifold is about. For example topological restrictions on the underlying space. An example of this kind would be the following theorem for Lorentzian manifolds and goes as follows:

Theorem: Let $(M,g)$ be a closed, simply connected Lorentz manifold and assume $M$ and $g$ are analytic. Then the isometry group of $M$ is compact.

Are there general results about closed Semi-Riemannian manifolds which have a non-compact isometry group?

Background: By a theorem of Myers and Steenrod the isometry group of a Riemannian manifold is a lie group. The same is true for Semi-Riemannian manifolds where the idea of a possible proof (which also works for Riemannian manifolds) is to embed the isometry group into the bundle of orthonormal frames as a closed submanifold. In the case of Riemannian manifolds this immediately yields that the isometry group of a compact Riemannian manifold has to be compact. But this is in general false for Semi-Riemannian manifolds since the bundle of orthonormal frames is not necessarily compact in this case even if the manifold is. So I ask myself if there are general statements about the phenomenon of a closed Semi-Riemannian manifold with non-compact isometry group. One statement that I know of is about Lorentzian manifolds and goes as follows:

Theorem: Let $(M,g)$ be a closed, simply connected Lorentz manifold and assume $M$ and $g$ are analytic. Then the isometry group of $M$ is compact.

Are there general results about closed Semi-Riemannian manifolds which have a non-compact isometry group?

Background: By a theorem of Myers and Steenrod the isometry group of a Riemannian manifold is a lie group. The same is true for Semi-Riemannian manifolds where the idea of a possible proof (which also works for Riemannian manifolds) is to embed the isometry group into the bundle of orthonormal frames as a closed submanifold. In the case of Riemannian manifolds this immediately yields that the isometry group of a compact Riemannian manifold has to be compact. But this is in general false for Semi-Riemannian manifolds since the bundle of orthonormal frames is not necessarily compact in this case even if the manifold is. So I ask myself if there are general statements about the phenomenon of a closed Semi-Riemannian manifold with non-compact isometry group.

Edit: My goal is to understand a bit better what the intuition behind compactness / non-compactness of the isometry of a closed Semi-Riemannian manifold is. For example topological restrictions on the underlying space. An example of this kind would be the following theorem for Lorentzian manifolds:

Theorem: Let $(M,g)$ be a closed, simply connected Lorentz manifold and assume $M$ and $g$ are analytic. Then the isometry group of $M$ is compact.

Source Link
JS.
  • 893
  • 6
  • 15

Closed Semi-Riemannian manifolds with non-compact isometry group

Are there general results about closed Semi-Riemannian manifolds which have a non-compact isometry group?

Background: By a theorem of Myers and Steenrod the isometry group of a Riemannian manifold is a lie group. The same is true for Semi-Riemannian manifolds where the idea of a possible proof (which also works for Riemannian manifolds) is to embed the isometry group into the bundle of orthonormal frames as a closed submanifold. In the case of Riemannian manifolds this immediately yields that the isometry group of a compact Riemannian manifold has to be compact. But this is in general false for Semi-Riemannian manifolds since the bundle of orthonormal frames is not necessarily compact in this case even if the manifold is. So I ask myself if there are general statements about the phenomenon of a closed Semi-Riemannian manifold with non-compact isometry group. One statement that I know of is about Lorentzian manifolds and goes as follows:

Theorem: Let $(M,g)$ be a closed, simply connected Lorentz manifold and assume $M$ and $g$ are analytic. Then the isometry group of $M$ is compact.