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Isotropy in torus-topology

  1. Aug 30, 2010 #1

    tom.stoer

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    My question is if a torus admits coordinates which guarantuee isotropy.

    Background:

    In cosmology one usually assumes homogenity and isotropy of the universe. These principles are respected by most prominent cosmological models. Now what about the torus universe? Is there a choice of coordinates for which isotropy is guarantueed?

    Using standard torus coordinates via embedding obviously violates isotropy. Using a square with opposite edges identified and cartesian coordinates on this square violates isotropy as well.

    My feeling is that isotropy is always violated, i.e. that the torus topology does not allow for a geometry which respects isotropy. My reasoning goes as follows: using the square (cube, ...) one immediately sees that for a straight curve parallel to the edges of the square the curve always closes with winding number 1. A curve not parallel to the edges will close with winding number >1 (in rational cases) or it will never close (in irrational cases). But of course this is only one counter example for one specific geometry, not a general proof.

    So my question is if there is a geometry on a torus which respects isotropy.
     
    Last edited: Aug 30, 2010
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  3. Aug 30, 2010 #2

    Office_Shredder

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    In general, how is isotropy defined? Do you have to follow geodesics?
     
  4. Aug 30, 2010 #3

    tom.stoer

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    Good question.

    The general statement is that "all directions are equivalent". I would suggest to define "directions" via geodesics (= generalized straight lines which are equivalent to shortest connection of twopoints in the absence of torsion; in the presence of torsion the equivalence breaks down).
     
  5. Aug 30, 2010 #4

    tom.stoer

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    This what Wikipedia says:

    "In mathematics, an isotropic manifold is a manifold in which the geometry doesn't depend on directions. A simple example is the surface of a sphere.

    ...

    A homogeneous space can be non-isotropic (for example, a flat torus !!!), in the sense that an invariant metric tensor on a homogeneous space may not be isotropic."

    So according to Wikipedia (w/o proof) flat geometry on a torus is not compatible with isotropy.
     
  6. Aug 30, 2010 #5

    lavinia

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    the Wikipedia link is a stub and explains nothing. Can you define isotropic in rigorous mathematical terms?
     
    Last edited: Aug 30, 2010
  7. Aug 30, 2010 #6

    tom.stoer

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    A space is called homogenous if its properties do not depend on the location. A space is called isotropic if (at each point) its properties do not depend on the direction (one looks).

    I have no rigorous mathematical definition, but the example of the torus should make it clear: a torus with flat geometry is definitly homogeneous, i.e. its geometric properties do not depend on the location (it looks flat everywhere). In addition it looks isotropic locally. But globally its properties do depend on the direction. Its properties do not change along a straight curve, but there are curves with different winding properties depending on the direction of the curve.

    Perhaps a thesaurus may help: isotropy - the property of being isotropic; having the same value when measured in different directions.
     
  8. Aug 30, 2010 #7

    lavinia

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    the Wikipepedia article is a stub. Can you define isotropic rigorously?
     
  9. Aug 30, 2010 #8

    tom.stoer

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    I saw that! As I said I have no rigorous definition. You can check the following link, perhaps it becomes clearer.

    http://www.answers.com/topic/isotropy-3

    "Within mathematics, isotropy has a few different meanings:
    (1) Isotropic manifolds: Some manifolds are isotropic, meaning that the geometry on the manifold is the same regardless of direction. A similar concept is homogeneity. A manifold can be homogeneous without being isotropic. But if it is inhomogeneous, it is necessarily anisotropic."
     
  10. Aug 31, 2010 #9

    tom.stoer

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    I guess I have both the definition of isotropy and the explanation why the torus fails to be isotropic.

    A friend of mine found the following definition (Lee: "Riemann Manifolds")

    "..given a point p \in M, M is isotropic at p if there exists a Lie group G acting smoothly on M by isometries such that the isometry subgroup G_p \subset G acts transitively on the set of unit vectors in T_p M."

    I think that it's is clear what it means: given any p and two unit vectors x and y in T_M(p) one can find a rotation g which sends x to y, g(x) = y. Therefore the action is transitive locally i.e. for each p. But a rotation on the torus cannot be extended globally in a smooth manner. This becomes intuitively clear using a quadratic chart with cartesian coordinates covering nearly all of the (flat) 2-torus = leaving out only a small strip along the edges; within the square everything is fine, but in the small strip one has to "unwind" the rotation (which breaks local flatness of the torus and causes anisotropy within the strip) or one has to abandon smoothness at all.

    So the crucial fact is GLOBAL transitivity and smoothness.
     
  11. Aug 31, 2010 #10

    lavinia

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    So projective space is isotropic.

    the set of lattice preserving isometries of Euclidean space is what?

    I guess it is just the group of invertible integer matrices of determinant +-1
     
    Last edited: Aug 31, 2010
  12. Aug 31, 2010 #11

    lavinia

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    such spaces must have constant curvature - so which hyperbolic manifolds if any are isotropic?
     
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