# Contraction of mixed tensor

• I

## Main Question or Discussion Point

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

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robphy
Homework Helper
Gold Member
For a general $A^{mn}{}_{kl}$, no.

haushofer
It's only true when your tensor is symmetric in both upper and lower indices.

robphy
Homework Helper
Gold Member
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,
$A^{mn}{}_{lp}=\frac{1}{2}\left( A^{mn}{}_{lp} + A^{nm}{}_{pl} \right)$.

haushofer