Contraction of mixed tensor

  • #1
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Main Question or Discussion Point

Is that true in general and why:
$$A^{mn}_{.~.~lm}=A^{nm}_{.~.~ml}$$
 

Answers and Replies

  • #2
robphy
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For a general ##A^{mn}{}_{kl}##, no.
 
  • #3
haushofer
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It's only true when your tensor is symmetric in both upper and lower indices.
 
  • #4
robphy
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It's only true when your tensor is symmetric in both upper and lower indices.
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)##.
 
  • #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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