# I Contraction of mixed tensor

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1. Dec 11, 2016

### arpon

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

2. Dec 11, 2016

### robphy

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

3. Dec 12, 2016

### haushofer

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

4. Dec 12, 2016

### robphy

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. Dec 12, 2016

### haushofer

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