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Question: Is there any (absolute) geometry which is not "infinitesimally Euclidean"?

Context: All of the geometries listed on the Wikipedia page "Foundations of Geometry" (describing axiomatic formulations of geometry) seem to correspond to special cases of absolute geometry, and it seems like any absolute geometry is either hyperbolic, elliptic, or Euclidean (parabolic?) according to the version of the parallel postulate used, perhaps equivalently according to the type of curvature of the underlying geometric space. These all seem to have realizations or models as Riemannian manifolds of some sort.

Any geometry of (smooth) manifolds seems to be infinitesimally Euclidean, even for those without a Riemannian metric, since each neighborhood is (diffeomorphic) homeomorphic to Euclidean space.

Hyperbolic geometry seems to be the study of Riemannian manifolds with negative curvature, elliptic geometry the study of Riemannian manifolds with positive curvature, and Euclidean geometry the special case where there is no curvature. But obviously every neighborhood of a Riemannian manifold is diffeomorphic Euclidean space, thus even the Riemannian geometry of spaces like the torus, which is neither strictly elliptic nor hyperbolic, is infinitesimally Euclidean.

Thus it seems like to me that all of the elementary geometric axioms determine every aspect of the geometric space (e.g. that it must be a Riemannian manifold) except the curvature -- thus changes in the parallel postulate seem to correspond to different values of the curvature of the space.

Am I understanding this correctly? I had thought previously that the term geometry could be applied to spaces so abstract that they could not be embedded in any Euclidean space, and in particular were not infinitesimally Euclidean, but now I am not so sure. Any clarification would be appreciated. The question stems in part from my reading of Agricola and Friedrich's "Elementary Geometry" (also of the original German version), so perhaps if you have read some of that book as well you might understand better the source of my misunderstanding.

Chill2Macht
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I know this is an old question, but I just discovered it, and I'm a little confused by the responses from others, because there is an obvious problem with the following claim you make:

Any geometry of (smooth) manifolds seems to be infinitesimally Euclidean, even for those without a Riemannian metric, since each neighborhood is (diffeomorphic) homeomorphic to Euclidean space.

You are confusing the category of smooth manifolds with the category of Riemannian manifolds, or the subject of differential topology with the subject of differential geometry.

More specifically, it makes no sense to talk about “geometry” in the category of mere smooth manifolds (even though we often learn about the apparatus of bare smooth manifolds in courses called “differential geometry”). The “geometry” is defined by the Riemannian metric and is something over and above the bare smooth structure. There is thus no sense at all to the statement that “the geometry of any smooth manifold is infinitesimally Euclidean”: there’s no geometry to speak of in the first place! Mere diffeomorphisms don’t capture, transport, or preserve geometry; isometries do.

So yes, every smooth manifold is locally diffeomorphic to Euclidean space, and yes this has nothing to do with whether or not the manifold is endowed with a metric structure. There are thus no local invariants in differential topology, for the reason you state: all objects in the category of smooth manifolds are locally diffeomorphic. But there are local invariants in differential geometry; not all objects in that category are locally isometric.

symplectomorphic
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    In particular, the geometry of the sphere, for example, is not "infinitesimally Euclidean." As you say, the sphere has positive curvature, but Euclidean space has zero curvature. The geometry around a point of zero curvature will be infinitesimally Euclidean, but not all points your torus, for example, have zero curvature: at those points the geometry won't be infinitesimally Euclidean. – symplectomorphic Mar 16 '17 at 00:56
  • I'm coming here even later, but I want to make sure I understand - the scalar curvature would count as a "local invariant" in Riemannian geometry, right? That's an example of something which does not exist in the smooth category, but can be used to differentiate Riemannian manifolds. So you could have a manifold, locally diffeomorphic to Euclidean space, but which can be given a metric non-isometric to flat Riemann, yes? – levitopher Feb 19 '21 at 03:23
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    @levitopher: yes. Any smooth manifold is locally diffeomorphic to Euclidean space. But that doesn’t mean their “geometry” is locally Euclidean. A diffeomorphism is far too general and wild to capture “geometry”: for that you need something stronger, a metric (which determines the local geometry) and the notion of local isometry (which says whether geometries are “the same”). Not all Riemannian manifolds — smooth manifolds equipped with the extra structure of a Riemannian metric — are flat. – symplectomorphic Feb 19 '21 at 03:37
  • I accidentally deleted another comment I had added in 2017, the spirit of which I restore here for posterity. Another looseness in the OP’s question is the conflation of “local” and “infinitesimal.” We usually distinguish them. Local properties hold in an open set around a point. Infinitesimal properties hold in the tangent space at a point. In this sense we can say Riemannian manifolds are infinitesimally but not necessarily locally Euclidean. (They are infinitesimally Euclidean because their tangent spaces are just inner product spaces, which are all isometric to Euclidean space.) – symplectomorphic Feb 19 '21 at 04:08
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Riemann in his famous essay of 1854 considered only metrics that are infinitesimal. For this reason Riemannian geometry properly speaking is only concerned with this type of manifold. A generalisation is known as Finsler spaces. Here infinitesimally the space looks like a Banach space, which is more general than Euclidean space.

Mikhail Katz
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  • You are right that "infinitesimal" in the sense of geometry probably has to mean tangent spaces of a smooth manifold. Thus your mention of Finsler spaces seems to be exactly on target -- from the first paragraph of the Wikipedia article on Finsler manifolds (https://en.wikipedia.org/wiki/Finsler_manifold) -- "Finsler manifolds non-trivially generalize Riemannian manifolds in the sense that they are not necessarily infinitesimally Euclidean. This means that the (asymmetric) norm on each tangent space is not necessarily induced by an inner product (metric tensor)." – Chill2Macht Jan 27 '17 at 21:32
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The subject of topology, and its sub-subject metric spaces, contains many examples of "absolute geometry" spaces that are not infinitesmally Euclidean, neither in the metric sense, nor the smooth sense, nor the topological sense. These subjects abound with many natural examples.

For example, infinite dimensional Hilbert spaces, which have many examples amongst function spaces, are not locally homeomorphic to Euclidean space. And yet their geometry is of intense interest. Take a look at any functional analysis book.

For another example, there is an entire theory devoted to metric spaces of nonpositive curvature. See the book of Bridson and Haefliger for a better feel.

Lee Mosher
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    Building on this, I've always wondered: are there natural examples of classes of manifold type objects which instead of being locally homeomorphic to Euclidean space are locally homeomorphic to some other kind of well-understood topological space? – Vik78 Oct 16 '16 at 14:12
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    Banach manifolds are examples. Universal Menger compacta, whose existence was established in Bestvina's thesis, are other examples. – Lee Mosher Oct 16 '16 at 14:13
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    Thanks. I always just felt like the definition of manifold could easily be generalized using any topological space, and wondered why this approach wasn't more common. – Vik78 Oct 16 '16 at 14:15
  • There's also matchbox manifolds, which are locally cantor spaces (or the product with a Euclidean ball). Solenoids fall into this category as some of the simplest examples. – Dan Rust Oct 16 '16 at 14:19
  • @Vik78: Part of the trouble with such generalizations is that manifolds have a "homogeneity" property which is hard to mimic in situations that are not locally Euclidean. That, perhaps, might explain why it took so long for the theory of universal Menger compacta to be put on solid ground. But local homogeneity is evident for Hilbert spaces and other normed vector spaces, which perhaps explains why they have been around for much longer. – Lee Mosher Oct 16 '16 at 14:31
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Consider the geometry of the surface of a cone. The curvature is singular in one direction around the bottom edge, and in every direction at the tip. Thus the geometry is not even approximately Euclidean at those points. For a space that is not approximately Euclidean anywhere, you're getting into the domain of fractals, like would be formed by some higher dimension analogue of the Weierstrauss function. I'm unsure of whether the geometry of such a complicated space could be axiomatized into an absolute geometry. It is, I think, worth a look.

Sean Lake
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Geometry is what mathematicians call geometry.

Absolute plane geometry, referring to the theory you've linked, is not an attempt at being a general theory of geometry — instead, it axiomatizes an extremely rigid structure that encompasses exactly two models: the euclidean plane and the hyperbolic plane.

To the best of my knowledge, the only real reasons for the theory is:

  • to provide the setting in which one can consider the question "can you prove the parallel postulate?"
  • a starting point for learning hyperbolic geometry