- #1
"pi"mp
- 129
- 1
I've been thinking about the number of degrees of freedom in a tensor with n indices in 2-dimensions which is traceless and symmetric. Initially, there are [itex] 2^{n} [/itex] degrees of freedom. The hypothesis of symmetry provides n!-1 number of conditions of the form:
[tex] T_{i_{1}, \ldots i_{n}}- T_{\sigma i_{1}, \ldots \sigma i_{n}} =0 [/tex]
since there are n! permutations, including identity. So why doesn't this reduce the number of degrees of freedom by n!-1 ? Are these conditions not all independent?
I think I'm correct that tracelessness reduces the degrees of freedom by [itex] \binom{n}{2} [/itex] since we choose two indices to contract with [itex] g^{ab} [/itex], but maybe not.
[tex] T_{i_{1}, \ldots i_{n}}- T_{\sigma i_{1}, \ldots \sigma i_{n}} =0 [/tex]
since there are n! permutations, including identity. So why doesn't this reduce the number of degrees of freedom by n!-1 ? Are these conditions not all independent?
I think I'm correct that tracelessness reduces the degrees of freedom by [itex] \binom{n}{2} [/itex] since we choose two indices to contract with [itex] g^{ab} [/itex], but maybe not.