Let $H$ be a Hilbert space, and let
$$H_n = \otimes_n H = \Big\{\sum_{i_1,\ldots, i_n} \alpha_{i_1, \ldots, i_n} \big(e_{i_1} \otimes \cdots \otimes e_{i_n}\big) : \sum_{i_1,\ldots, i_n} |\alpha_{i_1, \ldots, i_n}|^2<\infty \Big\}$$
denote the $n$-fold tensor product of $H$. Here $\{e_i\}$ denotes a basis element for $H_i$ and $\alpha_{i_1, \ldots, i_n} \in \mathbb{C}$. Please correct me if this definition is incorrect.
I am trying to understand a particular subset of $H_n$, namely the symmetric tensor product. This is defined as the space
$$H_n^s = \Big\{\sum_{i_1,\ldots, i_n} \alpha_{i_1, \ldots, i_n} \big(e_{i_1} \otimes \cdots \otimes e_{i_n}\big) \in H_n: ~\alpha_{i_1, \ldots, i_n} = \alpha_{i_{\sigma(1)}, \ldots, i_{\sigma(n)}}~ \forall \sigma \in S_n,\Big\}$$
where $S_n$ is of course the permutation group of $n$ objects.
To me, this says that for every vector in $H_n^s$ if we swap the order of the coefficient's components then the result is the same coefficient we started with. However this is clearly wrong, as this would imply that the coefficients must all be the same. If anyone can provide a detailed breakdown of this definition that would be much appreciated.
Best Answer
The symmetric tensors are all symmetrized tensor products of vectors: $$ H^s_n = \left\{\sum_{\sigma \in S_n}v_{\sigma(1)}\otimes\cdots\otimes v_{\sigma(n)} \;:\; v_1,\dotsc,v_n \in H\right\}. $$ It follows that each basis element of $H^s_n$ corresponds to a selection of integers $1 \leq i_1 \leq i_2 \leq \cdots \leq i_n$, and that basis element is $$ \sum_{\sigma\in S_n}e_{i_{\sigma(1)}}\otimes\cdots\otimes e_{i_{\sigma(n)}}. $$ (Of course, you could normalize however you please.) This means we can write $$ H^s_n = \left\{\sum_{1\leq i_1\leq\cdots\leq i_n}a_{i_1,\cdots, i_n}\sum_{\sigma\in S_n}e_{i_{\sigma(1)}}\otimes\cdots\otimes e_{i_{\sigma(n)}} \;:\; a_{i_1,\cdots,i_n} \in \mathbb C\right\} $$ In terms of coordinates, this means that every $e_{i_1}\otimes\cdots\otimes e_{i_n}$ for any selection of $i_1,\dots, i_n$ must have the same coefficient as $e_{i_{\sigma(1)}}\otimes\cdots\otimes e_{i_{\sigma(n)}}$ for any $\sigma \in S_n$. Hence we could also write $$ H^s_n = \left\{\sum_{i_1,\cdots,i_n} a_{i_1,\cdots,i_n}e_{i_1}\otimes\cdots\otimes e_{i_n} \;:\; \forall\sigma\in S_n.\: a_{i_{\sigma(1)},\cdots,i_{\sigma(n)}} = a_{i_1,\cdots,i_n}\right\}. $$ This says that if any two coefficients $a_{i_1,\cdots,i_n}$ and $a_{i'_1,\cdots,i'_n}$ draw their indices from the same multiset, then $a_{i_1,\cdots,i_n} = a_{i'_1,\cdots,i'_n}$; it does not say anything about how e.g. $a_{1,2,3}$ and $a_{1,1,3}$ are related.