Is the Contraction of a Mixed Tensor Always Symmetric?

[arpon]

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

 

[robphy]

For a general $A^{mn}{}_{kl}$, no.

 

[haushofer]

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

 

[robphy]

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]

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