Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Levi-civita connection assumptions

  1. Apr 11, 2010 #1
    In a lot of textbooks on relativity the Levi-Civita connection is derived like this:

    [tex]V=V^ie_i [/tex]
    [tex]dV=dV^ie_i+V^ide_i [/tex]
    [tex]dV=\partial_jV^ie_idx^j+V^i \Gamma^{j}_{ir}e_j dx^r [/tex]
    which after relabeling indices:
    [tex]dV=(\partial_jV^i+V^k \Gamma^{i}_{kj})e_i dx^j [/tex]

    so that the covariant derivative is defined as:

    [tex]\nabla_j V^i=\partial_jV^i+V^k \Gamma^{i}_{kj}[/tex]

    However, the connection coefficient [tex]\Gamma^{i}_{kj} [/tex] is torsion-free by definition, as [tex]de_i=\Gamma^{j}_{ir}e_j dx^r [/tex] implies that
    (1) [tex]\partial_r e_i=\Gamma^{j}_{ir}e_j [/tex].

    If [tex]e_i=\partial_i[/tex] then since [tex]\partial_i\partial_r=\partial_r\partial_i[/tex] then by (1):

    [tex]\Gamma^{j}_{ir}e_j=\Gamma^{j}_{ri}e_j [/tex]

    or that the bottom two indices are symmetric which is the torsion-free condition.

    I have two questions. Is the equation [tex]\partial_r e_i=\Gamma^{j}_{ir}e_j [/tex] true in general for any connection? And also, where did the torsion-free assumption enter into the derivation above?
     
  2. jcsd
  3. Apr 16, 2010 #2
    Hello,

    I'm not sure about the derivation you gave, I'm a bit hazy on all this. As far as I know the Levi-Civita connection is, by definition, the (unique) connection which is compatible with the metric and has zero torsion. (Perhaps this is shown in Jost, Differential Geometry or similar.)

    Presumably this means that if you remove the zero torsion hypothesis, the connection is no longer uniquely defined (and it is not the L-C connection).

    Dave
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook