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I Contraction of mixed tensor

  1. Dec 11, 2016 #1
    Is that true in general and why:
    $$A^{mn}_{.~.~lm}=A^{nm}_{.~.~ml}$$
     
  2. jcsd
  3. Dec 11, 2016 #2

    robphy

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    For a general ##A^{mn}{}_{kl}##, no.
     
  4. Dec 12, 2016 #3

    haushofer

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    It's only true when your tensor is symmetric in both upper and lower indices.
     
  5. Dec 12, 2016 #4

    robphy

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    That's the requirement that
    ##A^{mn}{}_{lp}=A^{(mn)}{}_{(lp)}=\frac{1}{4}\left( A^{mn}{}_{lp} + A^{nm}{}_{lp} +A^{mn}{}_{pl}+A^{nm}{}_{pl} \right)##.
    But I think that's too strong.

    From what was given,
    I think that [if I'm not mistaken] the only requirement is that ##A^{mn}{}_{lp}=A^{nm}{}_{pl}## (simultaneous swap),
    that is,
    ##A^{mn}{}_{lp}=\frac{1}{2}\left( A^{mn}{}_{lp} + A^{nm}{}_{pl} \right)##.
     
  6. Dec 12, 2016 #5

    haushofer

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    Yes, you're right. The tensor is also allowed to be antisymmetric in both upper and lower indices.
     
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