Quick question to clear up some confusion on Riemann tensor and contraction

  1. May 2, 2011 #1
    Let's say I want to calculate the Ricci tensor, [tex]R_{bd}[/tex], in terms of the contractions of the Riemann tensor, [tex]{R^a}_{bcd}[/tex]. There are two definitions of the Riemann tensor I have, one where the [tex]a[/tex] is lowered and one where it is not, as above.

    To change between the two all that I have ever seen written is 'we lower the indices' but I don't think I fully understand this. Does this mean...

    [tex]R_{abcd} = g_{aa} {R^a}_{bcd}[/tex]

    So the answer to my original question of finding the Ricci tensor is...

    [tex]R_{bd} = g^{ac} g_{aa} {R^a}_{bcd}[/tex]

    Following this, I also have written before me that...

    [tex]{R^b}_{bcd} = 0[/tex] since [tex]R_{abcd}[/tex] is symmetric on a and b. Shouldn't this be antisymmetric on a and b?

    Sorry if these are basic questions but I'm finding the vagueness of 'lowering the indices' a bit confusing...

  2. jcsd
  3. May 2, 2011 #2


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    You can't have the same index appearing 3 times in an expression, what you mean is

    [tex]R_{abcd} = g_{ae} {R^e}_{bcd}[/tex]​
  4. May 2, 2011 #3
    The Riemann tensor


    The raising and lowing of indices is done using the fundamental metric tensor [itex]g_{\alpha\beta}[/itex] and the inverse metric [itex]g^{\alpha\beta}[/itex].

    The Ricci tensor

    [tex]R^{\alpha}_{\;\beta\gamma\delta}\rightarrow R^{\alpha}_{\;\beta\alpha\delta}=R_{\beta\delta}[/tex]

    is a contraction of the first and third indices.
  5. May 2, 2011 #4


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    I would argue that [tex]
    is the more fundamental expression for the Riemann tensor
    since this is defined from the derivative operator [without reference to a metric].
    Its trace yields the Ricci tensor [as jfy4 wrote]... and this too doesn't make use of a metric.

    To lower the upper index of Riemann or to get the scalar-curvature from Ricci now requires a metric.
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