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Riemannian Geometry , Levi-Civita connection, metric form q

  1. Jan 16, 2015 #1
    Conventional GR is based on the Levi-Civita connection.

    From the fundamental theorem of Riemann geometry - that the metric tensor is covariantly constant, subject to the metric being symmetric, non-degenerate, and differential, and the connection associated is unique and torsion-free - the connection can be determined by the metric from a relativity simple formula.

    BUT , doesn't whether the metric is symmetric or not, depend upon the choice of coordinates when specifying the metric. E.g- Schwarzschild metric given in spherical polar coordinates is diagonal, and so symmetric.
    However, the extended Schwarzschild metric in Eddington-Finkelstein coordinates is not diagonal, contains a dudr term, but not a drdu term. Does this mean that for the metric in Eddington-Finkelstein coordinates , the connection can no longer be simply computed from the metric?

    Thanks in advance.
  2. jcsd
  3. Jan 16, 2015 #2


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    The metric is always symmetric. This is in the definition of a metric.

    When people write out the metric in the familiar ##ds^2=-dt^2+...## form, they are actually writing out the contraction ##ds^2=g_{ij}dx^i dx^j##. But since ##dx^idx^j=dx^jdx^i## the ##g_{ij}## and ##g_{ji}## terms just get added together so you don't see both terms, just the combination ##g_{ij}+g_{ji}=2g_{ij}##.
  4. Jan 17, 2015 #3
    Is this true for any geometry and not just Riemannian?
  5. Jan 17, 2015 #4


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    If the metric were not symmetric, then the dot product of two vectors would not be commutative.
  6. Jan 17, 2015 #5


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    Mathematically, inner products are symmetric by definition. In a coordinate system you are merely expressing the inner product ( in fact any tensor) in terms of the coordinates. You are not changing it in any way.
  7. Jan 18, 2015 #6


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    It would be anti-commutative , like Grassmann variables...
    But wouldn't that make a very weird universe?? It blows my mind to think that [itex] r \theta = - \theta r [/itex]
  8. Jan 18, 2015 #7


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    It wouldn't have to be "anti-commutative" unless it were totally anti-symmetric. A general tensor will have both symmetric and anti-symmetric parts. Usually the term "anti-commutative" means that variables anti-commute ##\{a,b\}=0##.

    Anyways, there is a structure that IS anti-commutative that sometimes we use for manifolds, and that is called a symplectic structure. The basic quantity of interest there; however, is not called a metric, but the symplectic form. We use this for, e.g. geometric Hamiltonian mechanics.
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