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Weyl tensor

  1. May 26, 2010 #1
    I found the formula for the number of independent components of
    Weyl tensor in n-dimensional manifold:
    [tex]
    (N+1)N/2 - \binom{n}{4} - n(n+1)/2~~~~~N=(n-1)n/2
    [/tex]
    This expression implies that in 3 dimension Weyl tensor has 0 independent
    components, so it's 0. Does it implies that any three-dimensional manifold
    is conformally flat (maybe the formula I've written above is incorrect for n<4)?
     
  2. jcsd
  3. May 26, 2010 #2

    bcrowell

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    WP has some infromation that seems relevant: http://en.wikipedia.org/wiki/Weyl_tensor
    Physically, I would associate the Weyl tensor with the existence of tidal forces, which conserve the spatial volume of a free-falling cloud of particles. In 2+1 dimensions, this can exist. In 1+1, it can't. From the WP article, it sounds like maybe the Cotton tensor, rather than the Weyl tensor, is the appropriate way to measure tidal forces in 2+1...?
     
  4. May 26, 2010 #3
    Thanks for answer. The vanishing of Weyl tensor is nessesary condition but not sufficient
    for a metric to be conformally flat. In dimension >=4 it turns out to be also sufficient
    condition but in lower dimensional manifold it is not.
    In Wiki article http://en.wikipedia.org/wiki/Weyl_tensor#Conformal_rescaling
    about Weyl tensor there is said that metric is conformally flat if it fullfil some PDE
    and in dimension >=4 the only integrability condition for this PDE is Weyl tensor=0.
    Maybe someone knows what this PDE has in common with conformal flatness.
    I have no idea, but it looks interesting.
    I'm also looking for simple prove that conformal mapping doesn't change Weyl tensor
    (I would love to find geometrical argument, but probably it's imposibble).
    Thanks for help
     
  5. May 26, 2010 #4
    There is no other alternative to the proof of this theorem. Actually it is all clear that a conformal transformation of metric tensor, i.e. [tex]\bar{g}_{ab}=e^{2\lambda}{g}_{ab},[/tex] with [tex]\lambda[/tex] being a scalar function of coordinates [tex]x^{a}[/tex] using a direct calculation, including the expansion of Christoffel symbols and Riemann tensor in the new coordinates, leaves the Weyl tensor unchanged and this all doesn't exceed two ordinary textbook pages.

    See e.g. Riemannian Geometry, E. P. Eisenhart, Princeton Press 1949, pp 89-90.

    There's already a formula to calculate the number of the independent components of the WT and it is

    [tex]\frac{1}{12}n(n+1)(n+2)(n-3)[/tex]

    which gives 0 for n=3 and obviously works fine for n's greater than 2. I don't know how you manage to derive your formula but if it works for any given n, you should first consult an expert and then think of a suitable journal to publish it in (though I for one am sure it has nothing new to tell but as a tacit work can be good.)

    Nupe. Any three dimensional pseudo-Riemannian* manifold is conformally flat iff the third order http://en.wikipedia.org/wiki/Cotton_tensor" [Broken] vanishes.

    * (this must be pseudo-Riemannian Weyl tensor is a measure of the pseudo-Riemannian manifolds)

    AB
     
    Last edited by a moderator: May 4, 2017
  6. May 26, 2010 #5
    Thanks for your answer.
    My formula is completly equivalent to yours - I've factorised mine and got the same result.

    Maybe you know why in dimension >=4 metric is confomally flat iff Weyl tensor=0
     
  7. May 26, 2010 #6
    See e.g. Riemannian Geometry, E. P. Eisenhart, Princeton Press 1949, pp 91-92.

    If you hit anything difficult to catch on, simply ask it here!

    AB
     
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