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Topology of the FLRW model

  1. Dec 10, 2014 #1
    So, I was trying to do a derivation of my own for the FLRW metric, since I couldn't understand the one Wald had. The spatial slice [itex]M[/itex] is a connected Riemannian manifold which is everywhere isotropic. That is, in every point [itex]p\in M[/itex] and unit vectors in [itex]v_1,v_2\in T_p\left(M\right)[/itex] there is an isometry [itex]\phi:M\rightarrow M[/itex] with [itex]\phi_*v_1 = v_2[/itex]. By isotropy, the sectional curvature is a pointwise function and by Schur's Lemma, [itex]M[/itex] it's constant.

    My question is the following. Under those assumptions, can it be proved that [itex]M[/itex] is simply connected and complete? The FLRW metric suggests that [itex]M[/itex] is:
    (a) a sphere if the curvature is constant.
    (b) a hyberboloid if the curvature is negative,
    (c) flat if the curvature is zero.
    all if which are connected, simply connected and complete manifolds.

    P.S. This may be the wrong section for this thread. If so I apologize :)
  2. jcsd
  3. Dec 10, 2014 #2

    Ben Niehoff

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    Statements about curvature are only local. There is a lot you can still do with the global topology. Most notably, you can take quotients by discrete subsets of the isometry group.

    For example, if M is positively curved, you can obtain lens spaces. If M has zero curvature, you can obtain tori. If M has constant negative curvature, there are loads of different things you can do, of which this is one example:


    So no, you cannot conclude that M is simply connected.

    However, you might be able to prove something if your space is globally isotropic rather than merely locally so. That is, if the isometry in your definition is required to be a global one. In the above examples, since you've quotiented out by a discrete group of global isometries, you will probably not have enough global isometries left over to map any unit vector into any other. For example, on a torus the rotational isometries that fix a point are reduced to a discrete group.
    Last edited: Dec 10, 2014
  4. Dec 10, 2014 #3


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    - Based on what Ben said I think that the real projective spaces of constant curvature are examples of a non-simply connected isotropic manifolds. Their fundamental group is Z/2Z.

    - No flat Riemannian manifold except Euclidean space can be globally isotropic. The same point that Ben made about the torus also applies. All such manifolds are quotients of flat Euclidean space by a discrete group that contains a lattice and arbitrary rotations of Euclidean space will not preserve a lattice. So for instance a flat cylinder or a Klein bottle can not be globally isotropic
    Last edited: Dec 10, 2014
  5. Dec 10, 2014 #4
    Thank you guys, that was really helpful! :)
  6. Dec 10, 2014 #5

    George Jones

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    You might like "The Shape and Topology of the Universe" by Luminet,

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