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Projection of the Riemann tensor formula.

  1. Nov 18, 2013 #1
    Suppose we are given two projection operators H' and H'' such that H' + H'' = 1, i.e. that any vector can be written as V = V' + V'' = (H' + H'') V. I'm trying to prove the formula

    $$R(X',Y'')Z' \cdot V'' = (Z' \cdot (\nabla'_{X'}B') + \left<X'\cdot B', Z' \cdot B'\right>)(Y'', V'') + (V'' \cdot (\nabla'_{Y''}B') + \left<Y''\cdot B'', V'' \cdot B''\right>)(X', Z')$$

    where ##B''(X',Y') = - \nabla''_{X'}Y'## and ##B'(Y'',V'') = - \nabla'_{Y''}V''##,

    $$(\nabla'_{X'}B')(Y'',V'') = \nabla'_{X'}(B'(Y'',V'')) - B'(\nabla'_{X'}Y'',V'') - B'(X', \nabla'_{X'}V'') $$,

    $$\left<X'\cdot B', Z' \cdot B'\right>)(Y'', V'') = \left<X'\cdot B'(Y'',.), Z' \cdot B'(.,V'')\right>$$

    where the latter defines the scalar product between the two one-forms ##\alpha=X'\cdot B'(Y'',.)## and ##\beta = Z' \cdot B'(.,V'')##, with

    $$\left<X'\cdot B'(Y'',.), Z' \cdot B'(.,V'')\right> = \left<\alpha, \beta\right> = g^{-1}(\alpha,\beta).$$ The Riemann tensor R is given by the usual formula.

    Do you know of any references where the formula above is proven? Or can you suggest a way it might be proved? My biggest problem is making sence of the expression

    $$\left<X'\cdot B'(Y'',.), Z' \cdot B'(.,V'')\right>$$

    What would that correspond to in a form that will be more recognizable from the terms in the Riemann tensor?
  2. jcsd
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