# Mixed symmetry property and degrees of freedom

1. Jul 26, 2012

### sourena

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.

2. Jul 28, 2012

### MathematicalPhysicist

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

Last edited: Jul 28, 2012
3. Jul 28, 2012

### MathematicalPhysicist

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.

4. Aug 7, 2012

### sourena

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.

Last edited: Aug 7, 2012