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Extension of Thurston's Geometrization Conjecture to four dimensions?

  1. Apr 25, 2005 #1
    Hello all,

    I was wondering whether there might be anything similar to Thurston's Geometrization Conjecture, except for four manifolds instead of three manifolds. (Essentially, this conjecture (iiuc) states that it is always possible to take an arbitrary 3-manifold and break it up into smaller "pieces" of which there are eight types, each of which is pretty well understood -- see the explanation in mathworld [1].) I think that in two dimensions it is called the "uniformization theorem" [2], and ref [2] states that "It would be very important if similar results could be proved in higher dimensions." So I was wondering whether anyone has even *stated* the idea in dim=4?


    [1] http://mathworld.wolfram.com/ThurstonsGeometrizationConjecture.html

    [2] http://pdg.cecm.sfu.ca/~warp/java/uniform/node2.html
  2. jcsd
  3. Apr 25, 2005 #2


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    the discussions cited are not very precise, since they just use the word "geometry" without further explanation, but it appears from the analogy with 2 dimensions, that the meaning is something like riemannian geometry, i.e. metrics and curvature are involved.

    in dimension 2 a compact connected manifold can be given a metric of constant curvature, and that curvature is zero iff the manifold is a torus, and positive iff the manidfold is a sphere and negative iff the manifold is of genus at least 2.

    so apparently thurston is discussing the metric geometry of three manifolds. I do not know if such conjectures exist in higher dimensions, but the poincare conjecture, one of the corollaries of thurstons work it seems, already were proved in all higher dimewnsions, so for some questions, the three dimensional case was actually the hardest and the last one settled.

    Apparently this is not the case for metric questions. There are however other popints of view toward generalizing the 2 dimensional uniformization theorem, such the shafarevich conjecture studied recently by Kolla'r and then others. this asks what the complex geometry is for universal covering spaces of algebraic varieties, in particular when they are holomorphically convex.

    the analogy is that the 2 dimensional uniformization theorem follows from the fact that manifolds of geneus zero are spheres hence have well known positive curvature, while compact manifolds of genus zero are covered naturally by the plane which is flat, and complex riemann surfaces of genus at least 2 are covered by the unit disc, or upper half plane, which has hyperbolic geometry, i.e. negative curvature. these metrics and curvatures then descend to the compact manifolds they cover.

    hence understanding the universal cover sheds light on the manifold covered. in this regard higher dimensions are more difficult since the number of possible covering manifolds is (literally) infinitely greater.
    Last edited: Apr 25, 2005
  4. Apr 26, 2005 #3
    The one thing (and probably the most important thing) that is left out, is what Thurston means by a "geometry". In his book, he lays them out as such:

    A model geometry (G,X) is a manifold X with a Lie group G of diffeomorphisms of X such that
    1. X is connected and simply connected
    2. G acts transitively on X, with compact point stabilizers
    3. G is not contained in any larger group of diffeomorphisms of X satisfying #2 above, and
    4. there is at least one compact manifold modeled on (G,X), i.e. as noted by math wonk, this would be a compact manifold with X as a universal cover.

    A pretty straightforward definition, but I do not know if anyone has tried finding all of these in 4-dimensions. I do know that all homogeneous complex surfaces were classified a few years back. So, that's a start... However, there are plenty of 4-manifolds out there with no global complex structure, e.g. 1-dimensional quaternionic projective space...

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