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The book Introduction to Topology by C. Adams and R. Franzosa says :

There are only three three-dimensional geometries that are both homogeneous and isotropic: the Euclidean, spherical, and hyperbolic geometries.

It has been just written without giving even a little reason(s)/explanation, let alone a proof of it.

What is an easiest rigorous proof for the mentioned claim from the book?

EDIT - For the definitions of homogeneous and isotropic spaces, I quote it from the same book:

To be isotropic means that at each point the geometry appears to be the same in all directions around the point. There is no preferred identifiable direction. This is a property that Euclidean 3-space $E^3$ has, for instance. However, if we form a geometry by taking $S^2 \times E$, for example, that geometry has different behavior in different directions.

To be homogeneous means that locally the geometry of the space is the same. Given any two points in the space, there is an isometry (a distance-preserving homeomorphism) from a neighborhood of one point to a neighborhood of the other.

Thank you.

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  • How do you define geometry, homogeneous and isotropic? Thanks. –  Oct 24 '15 at 07:29
  • @JohnMa - I edited. Actually I've studied them in cosmology as well, but they fortunately have the same definitions in maths and physics. –  Oct 24 '15 at 07:37
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    It seems that your question is answered here. (See in particular p.74) –  Oct 24 '15 at 08:18

1 Answers1

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Given tangent 2-planes $P \subset T_p M, Q \subset T_q M$, there is an isometry $f$ taking $p$ to $q$; then using that $M$ is isotropic pick a local isometry taking $df_p(P^\perp)$ to $Q^\perp$; this takes $df_p(P)$ to $Q$. So we have a local isometry taking any tangent 2-plane to another. Thus the sectional curvature is constant. If $M$ is simply connected, it must be one of the three geometries you mention. However, $M$ needn't be simply connected; $M = \Bbb{RP}^n$ with the round metric is a homogeneous isotropic space, for instance. Maybe simply connected is demanded by your definition of 'geometry'.

  • This is a good answer but it seems like it is skipping some steps. Doesn't this (by defining inner products on the tangent spaces) a priori assume that the manifolds have a Riemannian metric/connection, when all one is given (seemingly) is that the manifolds are metrizable? Not even that they are smooth/differentiable? – Chill2Macht Jul 11 '17 at 14:14
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    @Chill2Macht You need a notion of tangent space to get a notion of direction to get a notion of isotropic. –  Jul 11 '17 at 14:15
  • That's actually a very interesting insight -- I need to remember it, I had not heard of or thought of it before. I hope that doesn't sound sarcastic, I do mean it. – Chill2Macht Jul 11 '17 at 14:17
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    Sure, thanks. It actually turns out to be a major sticking point in building a theory of topological manifolds that they don't have a good notion of tangent bundle. –  Jul 11 '17 at 14:20