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:(adsbygoogle = window.adsbygoogle || []).push({});

[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.

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# Counting Degrees of Freedom in Tensor

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