Let $\mathscr{H}$ be a Hilbert space with orthonormal basis $\{e_{1},e_{2},...\}$ and let $\sigma$ be a permutation of the set $\{1,...,n\}$ for some $n \ge 1$. Consider the following object: $$v = e_{i_{1}}\otimes \cdots \otimes e_{i_{n}}$$ Notice that the indices of $v$ may repeat. For each $i = 1,2,...$, let $n_{i}$ the number of times that the index $i$ appears in $v$.
On page 24 of these lecture notes, it is stated that:
$$\|\sum_{\sigma}e_{i_{\sigma(1)}}\otimes \cdots \otimes e_{i_{\sigma(n)}}\|^{2} = n! \prod_{i=1}^{\infty}n_{i}!$$
However, according to my calculations, this should be: $$\|\sum_{\sigma}e_{i_{\sigma(1)}}\otimes \cdots \otimes e_{i_{\sigma(n)}}\|^{2} = \frac{n!}{ \prod_{i=1}^{\infty}n_{i}!}$$ instead, because expanding $\|\cdot\|^{2}$ one would get: $$\sum_{\sigma,\rho}\delta_{i_{\sigma(1)},i_{\rho(1)}}\cdots \delta_{i_{\sigma(n)},i_{\rho(n)}}$$ where $\delta_{i,j}$ denotes the Kronecker delta. This would be equal to the number of permutations $\sigma,\rho$ which agree with $n_{1}$ repetitions of the $i=1$ index, $n_{2}$ repetitions of the $i=2$ index and so on. Am I doing something wrong or the result of the text is wrong?
The formule in the book is correct! Indeed, to count the number of permutations, you could start by choosing your $\sigma$ (you have $n!$ choices). After that you need to choose $\rho$ such that $i_{\sigma(j)} = i_{\rho(j)}$ for all $j$. So $\rho(1)$ you have $n_{i_{\sigma(1)}}$ choices possible, for the next one you have $n_{i_{\sigma(1)}} -1$ choices and so one and so forth. You will have $\prod_{i=1}^\infty n_i!$ permutations $\rho$.