Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

K=-1 closed space

  1. Feb 18, 2010 #1
    In my cosmology lectures they say that a negative curvature gives an infinite space but I was thinking what about the inside of a torus. Isn't that a closed space too???

    Cant any value of k apart from 0 result in a closed space??
  2. jcsd
  3. Feb 18, 2010 #2


    User Avatar
    Science Advisor
    Homework Helper

    Remember in cosmology though, we assume a homogeneous and isotropic universe, which results in just tress types of spaces (flat, sphere, hyperboloid).
  4. Feb 18, 2010 #3


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    There are theorems called local to global theorems that constrain the global topology for a given curvature: http://en.wikipedia.org/wiki/Riemannian_geometry#Local_to_global_theorems These theorems are not so strong that they give a 100% correspondence between local and global properties.

    As you've noted, a flat space can have various topologies, such as trivial, cylindrical, toroidal, Mobius... In the non-flat topologies, you get a preferred frame of reference, which is essentially the frame in which the circumference is maximized (as opposed to other frames which see the circumference as Lorentz-contracted).

    There are various exotic possibilities for the global topology of the universe:

    One thing I didn't understand properly until recently was that in models with nonzero cosmological constants, the spatial topology is not necessarily correlated with the existence of a Big Crunch.

    [EDIT] I don't pretend to understand the local to global theorems, and like the OP, I would be interested in hearing more about how this applies to spaces with nonvanishing curvature.

    In the case of 2-dimensional space with a positive-definite metric, I do think I understand the possibilities to some extent. In the flat case, you have Euclidean geometry, which can have a variety of topologies (trivial, cylindrical, toroidal, Mobius). In the positive-curvature case, you get elliptic geometry, and all models of elliptic geometry are closed; that is, you can start from the axioms of elliptic geometry and prove results like an upper bound on the area of any triangle. I would be interested to know whether the negative-curvature case (hyperbolic geometry) admits any topology other than the usual one; I suspect that it doesn't.

    Making the analogy with 3+1 dimensions, I would conjecture that you only get wiggle room on the spatial topology if the spatial curvature is zero.
    Last edited by a moderator: Apr 24, 2017
  5. Feb 19, 2010 #4

    George Jones

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    It possible to have spacetimes that have closed (i.e., compact) spatial sections that don't have positive spatial curvature, but, as nicksauce has noted, some symmetry of the standard cosmological models has to be relaxed. Typically, (local) spatial homogeneity is retained. See articles by Luminet,


    and Chapter 15, Spatially Homogeneous Universe Models, in the book Einstein's General Theory of Relativity With Modern Applications in Cosmology by Gron and Hervik.

    A 3-dimensional torus (a compact quotient of Euclidean 3-space) can be used as a particular model of a (locally) homogeneous, closed universe that has flat spatial sections, and a compact quotient of 3-dimensional hyperbolic space can be used as a particular model of a (locally) homogeneous, closed universe that has spatial sections with negative curvature.
    Last edited: Feb 19, 2010
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook