Undergrad Is the Contraction of a Mixed Tensor Always Symmetric?

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The contraction of a mixed tensor is not always symmetric unless specific conditions are met. It is only true when the tensor is symmetric in both upper and lower indices. The requirement for symmetry can be expressed mathematically, but a simultaneous swap of indices is sufficient for certain cases. Additionally, tensors can also be antisymmetric in both upper and lower indices. Thus, the symmetry of the contraction depends on the properties of the tensor involved.
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Is that true in general and why:
$$A^{mn}_{.~.~lm}=A^{nm}_{.~.~ml}$$
 
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For a general ##A^{mn}{}_{kl}##, no.
 
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It's only true when your tensor is symmetric in both upper and lower indices.
 
haushofer said:
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)##.
 
Yes, you're right. The tensor is also allowed to be antisymmetric in both upper and lower indices.
 
In an inertial frame of reference (IFR), there are two fixed points, A and B, which share an entangled state $$ \frac{1}{\sqrt{2}}(|0>_A|1>_B+|1>_A|0>_B) $$ At point A, a measurement is made. The state then collapses to $$ |a>_A|b>_B, \{a,b\}=\{0,1\} $$ We assume that A has the state ##|a>_A## and B has ##|b>_B## simultaneously, i.e., when their synchronized clocks both read time T However, in other inertial frames, due to the relativity of simultaneity, the moment when B has ##|b>_B##...

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