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Weyl tensor on 3-dimensional manifold

  1. Aug 7, 2006 #1
    Hello, I wish to show that on 3-dimensional manifolds, the weyl tensor vanishes.
    In other words, I want to show that the curvature tensor, the ricci tensor and curvature scalar hold the relation

    eq0009MP.gif

    Please, if anyone knows how I can prove this relation or refer to a place which proves the relation, I will be most grateful.

    Thanks in advance
     
  2. jcsd
  3. Aug 7, 2006 #2

    pervect

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  4. Aug 7, 2006 #3

    robphy

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    One should be able to count the number of algebraically independent components of the Riemann tensor in n-dimensions... then compare to the number for Ricci.
     
  5. Aug 8, 2006 #4
    Can you please help me doing the calculations?
    I know the final answer is that the number of independent components is
    (n^2(n^2-1))/12

    Can you please show me how to reach this result?
     
  6. Aug 8, 2006 #5

    pervect

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    I'll take a shot at doing the Riemann.

    Consider its symmetries: it's anti-symmetric in the first two indicies, so

    R_abcd = -R_bacd

    thus if a=b, we know that the Riemann is zero

    Similarly, IF the Riemann is derived from a metric

    R_abcd = -R_abdc, i.e. it's anti-symmetric in the last two indices.

    I think we need to assume that the Riemann is a Riemann derived from a metric....

    So far we have basically shown by symmetry that the Riemann must be

    R(u)(v), where u and v are anti-symmetric rank 2 tensors, aka two forms.

    How many 2-forms do we have in 3-d space? We have

    x^y, x^z, and y^z - a total of three. Let's call them p, q, and s

    The order of the two-forms doesn't matter because

    R_abcd = R_cdab (symmetry under exchange of front pair with back pair)

    So our possibilities are so far

    pp, pq, ps, qq, qs, ss

    That's 6, which is the right answer. I can see that there aren't any completely anti-symmetric terms to eliminate in only three dimensions, therfore

    R_[abcd]=0

    doesn't add any constraints. An open question:

    R_a[bcd]=0

    is the last remaining symmetry, it must also not contribute??
     
  7. Aug 9, 2006 #6

    samalkhaiat

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    Last edited: Aug 9, 2006
  8. Aug 10, 2006 #7

    samalkhaiat

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    Start with the identity;

    [tex]R_{ac}=3R_{ac}-R_{ac}-R_{ac}+Rg_{ac}- \frac{R}{2}(3g_{ac}-g_{ac})[/tex]

    then, rewrite it in 3D, i.e put;

    [tex]3=g_{a}^{a}= \delta_{a}^{a}[/tex]

    So,

    [tex]g^{bd}R_{abcd}=R_{ac}g_{b}^{b}-R_{bc}g_{a}^{b}-R_{ad}g_{c}^{d}+R_{bd}g^{bd}g_{ac}-(1/2)R(g_{ac}g_{b}^{b}-g_{bc}g_{a}^{b})[/tex]

    or;

    [tex]g^{bd}R_{abcd}=g^{bd}[R_{ac}g_{bd}-R_{ad}g_{bc}+R_{bd}g_{ac}-R_{bc}g_{ad}+(1/2)R(g_{ac}g_{bd}-g_{ad}g_{bc})][/tex]

    Now, in 3D, Reimann and Ricci have the same number of independent components (6 each). Therefore the Reimann tensor is determined completely by the Ricci tensor and the above equation gives;

    [tex]R_{abcd}=R_{ac}g_{bd}-R_{ad}g_{bc}+R_{bd}g_{ac}-R_{bc}g_{ad}-(1/2)R(g_{ac}g_{bd}-g_{ad}g_{bc})[/tex]


    regards

    sam
     
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