## mixed symmetry property and degrees of freedom

How can I calculate degrees of freedom of a rank (o,3) tensor, Aabc, that is mixed symmetry and antisymmetric in the first 2 indices? By mixed symmetry I mean this:
Aabc+Acab+Abca=0.
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 You have 3*3*3=27 options for $A_{ijk}$. Now by antisymmetry terms like A_{abc}=-A_{bac}; A_{cab}=-A_{acb}; A_{bca}=-A_{cba}; So only 27-3=24 terms are independent, now after the mixed symmetry we are left with: 24-1=23 (cause one term depends on the other two). So we are left with 5 dof. Hope I helped somehow. Edit: Obviously that terms with the same first two indices are zero, and we have 3*2=6 such terms, so we are left with 23-6=17 dof. After that we have terms like: A_{abb}=-A_{bab}; A_{acc}=-A_{cac} ; A_{cbb}=-A_{bcb}; A_{caa}=-A_{aca} ; A_{bcc}=-A_{cbc} ; A_{baa}=-A_{aba} Which means in the end: 17-6=11 dof. I hope I counted right this time. :-D
 BTW this raises a nice programming task of how to compute some arbitrary tensor of rank $0\choose n$ with the above constraints. Pitty I am not that good programmer.

## mixed symmetry property and degrees of freedom

For a rank (0,3) tensor, Aabc, without any constraint, degrees of freedom are 216, a,b,c = 0, ..., 6.

If this tensor is antisymmetric in the first 2 indices, degrees of freedom dicrease to 90.

If it is mixed symmetry, the number of constraint equations are:

$\frac{n(n-1)(n-2)}{3!}$, a,b,c=0, ..., n.

For our example n = 6 so, the number of constraint equations are 20 therefore degrees of freedom are 70.