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Commutation of metric tensor

  1. Feb 13, 2009 #1

    I am still having a hard time with tensors...
    The answer is probably obvious, but is it always the case (for an arbitrary metric tensor [tex]g_{\mu \nu}[/tex] that [tex]g_{ab}g_{cd}=g_{cd}g_{ab}[/tex] ?

    I was trying to find a formal proof for that, and was wondering if we could use the relations:
    (1) [tex]g_{ab}=\frac{\partial x^{c}}{\partial x^{a}} \frac{\partial x^{d}}{\partial x^{b}} g_{cd}[/tex]
    (2) [tex]g_{cd}=\frac{\partial x^{a}}{\partial x^{c}} \frac{\partial x^{b}}{\partial x^{d}} g_{ab}[/tex]

    And then multiply the left-hand side of (1) and (2) together and use the fact that the fractions of partial derivatives commute with the metric tensor and cancel each other to get:
    =(\frac{\partial x^{c}}{\partial x^{a}} \frac{\partial x^{d}}{\partial x^{b}} g_{cd}) (\frac{\partial x^{a}}{\partial x^{c}} \frac{\partial x^{b}}{\partial x^{d}} g_{ab})
    = \frac{\partial x^{c}}{\partial x^{a}}\frac{\partial x^{a}}{\partial x^{c}} \frac{\partial x^{d}}{\partial x^{b}} \frac{\partial x^{b}}{\partial x^{d}} g_{cd} g_{ab}
    = g_{cd}g_{ab}[/tex]

    Does that make sense ?!

    Thanks for your help...
  2. jcsd
  3. Feb 13, 2009 #2


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    [itex]g_{ab} [/itex] is just a number, and so is [itex]g_{cd}[/itex], and numbers commute.

    for instance, [itex]g_{02} = 0[/itex], [itex]g_{00} = 1 [/itex] (or [itex]g_{00} = -1 [/itex]) depending on which your original definition of metric is)
  4. Feb 13, 2009 #3
    Argh! Thanks a lot, again I mixed up "tensors" with "tensor components"...
    So the commutation relation I wrote holds for the tensor components of any tensor (not only the metric), doesn't it ?

    On the other hand, the tensor (objects) do not necessarily commute with each other, right ?
  5. Feb 13, 2009 #4


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    no the tensor objects can be non-commuting things like matricies, e.g.

    [tex]\sigma ^{\mu \nu} = \frac{i}{2}[\gamma ^\mu, \gamma^\nu][/tex]
    where the gamma's are dirac 4x4 matricies
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