Symmetric tensors as sum of powers I am looking for formulas for writing a basis element of $ Sym^k(H) $ as sum of elements of the form $ v^{\otimes k} $ where $ v\in H $. Here $ H $ is a hilbert space and by basis element I mean the image of elements of the form $ e_{i_1}\otimes e_{i_2}\otimes....\otimes e_{i_k} $ under the map $ H^{\otimes k}\to Sym^k(H) $ where $ \{e_i\}_{i\in \mathbb{N}} $ is a orthonormal basis of $ H $.
Its easy to prove that they can be written as sum of $ v^{\otimes k} $. And its easy to derive the expressions using roots of unity for small values $ k $. But any reference for where I can find explicit expressions for any $ k $ would be helpful.
 A: There are of course many ways to do that; I like the following formula, for $e_1,\ldots ,e_k\in H$ (I write the product in $Sym^k(H)$ as a usual product):
$$e_1\ldots e_k=\frac{(-1)^k}{k!} \sum_{I\subset [1,k]} (-1)^{\# I}(\sum_{i\in I}e_i)^k\ .$$
To prove it, take the term of degree $k$ in  the equality $$\prod (1-\exp (e_i))=1-\sum \exp (e_i)+\sum_{i<j}\exp (e_i+e_j)+\ldots $$
A: In effect you are asking how to write monomials as sums of powers (with scalar coefficients; linear combinations of powers). The expression given by @abx is very elegant, but uses $2^k-1$ terms for $e_1 \dotsm e_k$ (I'm counting each $k$th power as a term). This can be lowered to $2^{k-1}$, and generalized to other monomials.
As you say in your question, it is not difficult to derive explicit expressions. This was carried out in


*

*Buczyńska, et al, Waring decompositions of monomials, 2013 (MR3017012)

*Unpublished work by Gwyneth Whieldon (personal communication)


No doubt, many, many other people have done similar work in the last 200 or 300 years. (Surely, expanding monomials as sums of powers must have been child's play to people like Euler, Gauss, and Sylvester, and many others...!) The explicit expressions are not unique; I don't know if Whieldon's expressions are the same as in the other paper.
The expression for a simple product is hardly more complicated than the one given by @abx:
$$
  e_1 \dotsm e_k = \frac{1}{2^{k-1} k!} \sum_{\substack{\epsilon \in \{\pm1\}^k \\ \epsilon_1 = 1}} \left(\prod_{i=1}^k \epsilon_i\right) \left( \sum_{i=1}^k \epsilon_i x_i \right)^k
$$
Explicitly,
$$
\begin{gather}
  e_1 e_2 = \frac{1}{4} \Big( (e_1+e_2)^2 - (e_1-e_2)^2 \Big), \\
  e_1 e_2 e_3 = \frac{1}{24} \Big( (e_1+e_2+e_3)^3 - (e_1+e_2-e_3)^3 - (e_1-e_2+e_3)^3 + (e_1-e_2-e_3)^3 \Big),
\end{gather}
$$
and so on. As promised, this uses $2^{k-1}$ instead of $2^k-1$ terms. The proof can be a small exercise (or see the article above) but I'm not aware of any elegant thing like the identity @abx used.
Explicit expressions for any monomial, using a minimal number of terms, are given in the article above. They are not too complicated: for the monomial $e_1^{a_1} \dotsm e_k^{a_k}$, instead of coefficients $\pm 1$, we use coefficients given by $(a_i+1)$th roots of unity in front of each $e_i$, except that $e_1$ always just has $1$; each term's coefficient is the product of the roots of unity that appear.
