Degrees of freedom in construction of a symmetric tensor

Question

How many degrees of freedom does a symmetric tensor satisfying

$$S^{ij}_{kl} = S^{ji}_{kl} = S^{ij}_{lk} =S^{kl}_{ij}$$

have? The indices range from $1$ to $n$.

Answer 1

If we just had $S^{ij}_{kl} = S^{ji}_{kl} = S^{ij}_{lk}$, the answer would be $\Big(n(n+1)/2\Big)^2$, because the upper indices can either be the same [$n$ choices] or different [$\binom{n}2$ choices], and same for the lower. However, with the added relations $S^{ij}_{kl}=S^{kl}_{ij}$, we can switch the top and bottom. There are two possibilities; assuming WLOG that $i\le j$ and $k\le l$, there are $n(n+1)/2$ tensors where $(i,j)=(k,l)$, and there are $\binom{n(n+1)/2}{2}$ tensors where $(i,j)\neq (k,l)$. Therefore, the dimension is $$ \Big(n(n+1)/2\Big)\Big(n(n+1)/2 + 1\Big)/2. $$