Alternative base of a symmetric tensor product Let $V$ be a vector space with basis $ (e_{1}, \cdots , e_{d} )$. The $n$ fold symmetric tensor product $\operatorname{Sym}^n(V)\subset V^{\otimes n}$ is the subspace of symmetric tensors. It can be obtained as the image of the projection (on the symmetric space...)
$$ S:\left\lbrace \begin{aligned} V^{\otimes n}\quad & \longrightarrow \quad V^{\otimes n}\\ e_{i_1}\otimes \cdots \otimes e_{i_n} & \longmapsto \frac{1}{n!} \sum_{\sigma\in\mathfrak{S}_n} e_{i_{\sigma(1)}}\otimes \cdots \otimes e_{i_{\sigma(n)}}\end{aligned} \right. $$
(example: $n=3, d\geq 3$
$$\it S(e_{1}\otimes e_2 \otimes e_{3})= \frac{\big({\small e_{1}\otimes e_2 \otimes e_{3} + e_{2}\otimes e_3 \otimes e_{1} + e_{3}\otimes e_1 \otimes e_{2} + e_{1}\otimes e_3 \otimes e_{2} + e_{3}\otimes e_2 \otimes e_{1} + e_{2}\otimes e_1 \otimes e_{3}} \big)}{6} $$)
A basis of $\operatorname{Sym}^n(V)$ is given by  (cf. link p.33 in this question)
$$ \Big\lbrace S(e_{i_1}\otimes \cdots \otimes e_{i_n}),\ 1\leq i_1 \leq i_2 \leq \cdots \leq i_n \leq d \Big\rbrace \tag{1} \label{1}$$
so that (cf. this other post)
$$\operatorname{dim}\big(\operatorname{Sym}^n(V) \big) = { d+n-1 \choose n} \tag{2} \label{2}$$
Question: Show that $\big\lbrace \mathbf{x}\otimes \mathbf{x} \otimes \cdots \otimes \mathbf{x},\ \mathbf{x} \in V \big\rbrace $ is another generating set, or even better, give a basis consisting of vectors of this form.

My motivation for this question came from the sentence around equation (3.48) p.38 of this lecture notes on quantum field theory. A representation of $SU(2)$ is then (completely) defined on vectors of the form $\mathbf{x}\otimes \mathbf{x} \otimes \cdots \otimes \mathbf{x}$. In the notes, $V:= \mathbb{C}^2$ so that $d=2, n=2s$ in (\ref{2}), i.e. $\operatorname{dim}\big(\operatorname{Sym}^n(V) \big)=2s +1$
so additional keywords: symmetric tensor product, representation of $SU(2)$, spin.
 A: In fact, this question had already been addressed. Let me still give first my line of thoughts, and in a second part an erring, and finally an equivalent problem.


*

*I first to took the "finite difference" 
$$ (\mathbf{x}+\boldsymbol{\delta})\otimes (\mathbf{x}+\boldsymbol{\delta}) \otimes \cdots \otimes (\mathbf{x}+\boldsymbol{\delta})- \mathbf{x}\otimes \mathbf{x} \otimes \cdots \otimes \mathbf{x}= \sum_{k=1}^n {n\choose k} S\big(\boldsymbol{\delta}^{\otimes k} \otimes \mathbf{x}^{\otimes (n-k)} \big) $$ 
which imitates derivation ($f:x\mapsto x^n\ \Rightarrow f'(x)=nx^{n-1}$) but since the equivalent of the third derivative always vanishes, this procedure cannot produce the basic vectors of (1).

*the correct linear combination is an adaptation of the polarization formulae which we recall: it usually relates quadratic to bilinear forms and more generally, if $\alpha: V \to \mathbb{C}$ is such that $\forall\ \lambda \in \mathbb{C},\ \forall\ \mathbf{x}\in V,\ \alpha(\lambda \mathbf{x})= \lambda^n \alpha(\mathbf{x})$ then the $n^{\text{th}}$ derived form or defect
$$\Delta^n\alpha\ (\mathbf{x}_1, \mathbf{x}_2,\cdots , \mathbf{x}_n):= \frac{1}{n!}\sum_{1\leq i_1 < i_2 < \cdots < i_k \leq n} (-1)^{n-k} \alpha (\mathbf{x}_{i_1} + \mathbf{x}_{i_2} + \cdots + \mathbf{x}_{i_k}) \tag{Polar} \label{Polar}$$
is $n$-linear and symmetric. (I took it from Drapala,Vojtechovsky, (2.1) p.4. I'll write a proof later).

*In our problem, we should thus have (if the formula is correct)
$$\mathbf{x}_1 \otimes \mathbf{x}_2 \otimes \cdots \otimes \mathbf{x}_n = \frac{1}{n!}\sum_{1\leq i_1 < i_2 < \cdots < i_k \leq n} (-1)^{n-k} (\mathbf{x}_{i_1} + \mathbf{x}_{i_2} + \cdots + \mathbf{x}_{i_k})^{\otimes n} \tag{Sol} \label{Sol}$$
(then replace each $\mathbf{x}_i$ by a $e_{i_i}$ (confusing notation) that appears in the vectors of (1))


A misleading connection between (\ref{Polar}) and our initial problem is the "realization" of a tensor $T\in V^{\otimes n}$ as an $n$-linear map: $T: V^* \times V^* \times \cdots \times V^* \longrightarrow \mathbb{C} $ (cf. e.g. here, at least for finite dimensional spaces). For example $e_{1}\otimes e_2 \otimes \cdots \otimes e_{2}$ can be thought as
$$e_{1}\otimes e_2 \otimes \cdots \otimes e_{2}:\left\lbrace \begin{aligned} V^* \times V^* \times \cdots \times V^* & \longrightarrow \quad \mathbb{C}\\ (\lambda_{1}, \lambda_2, \cdots , \lambda_{n})\quad & \longmapsto \lambda_{1}(e_{1}) \lambda_2(e_2)  \cdots  \lambda_{n}(e_2) \end{aligned} \right.  $$
to which one associates the following homogeneous map of order $n$
$$\alpha: \left\lbrace \begin{aligned} V^* & \longrightarrow \quad \mathbb{C}\\ \lambda \enspace & \longmapsto \lambda(e_{1})\ \lambda(e_2)^{n-1} \end{aligned} \right.  \tag{$\alpha$} \label{alpha}$$
whose $n^{\text{th}}$-derived form should in principle be $e_{1}\otimes e_2 \otimes \cdots \otimes e_{2}$. The problem is that $\alpha$ is not of the form $\mathbf{x}\otimes \mathbf{x} \otimes \cdots \otimes \mathbf{x}$.

The question is analogous to the generalization of the "Gauss reduction" (no english article... the one used in Sylvester's law of inertia), i.e. expressing a general homogeneous polynomial of degree $n$
$$ P(x_1,x_2,\cdots, x_d)= \sum_{i=1}^d a_i x_i^n + \sum_{i\neq j} b_{i,j} x_i^{n-1} x_j + \sum_{i\neq j,k} c_{i,j,k} x_i^{n-2} x_j x_k + \cdots \tag{Poly}\label{Poly} $$
as a sum of $n^{\text{th}}$ power of linear forms, i.e. $\exists\ (\alpha_1,\cdots , \alpha_r)\in \mathbb{R}^r$ and $ (l_1,\cdots , l_r)$ linear maps s.t.
$$ P(x_1,x_2,\cdots, x_d)= \sum_{p=1}^r \alpha_p l_p(x_1,x_2,\cdots, x_d)^n \tag{nPower}\label{nPower}$$
(A somewhat formal correspondence with our problem is given by $$P\ \longleftrightarrow\ \sum_{i=1}^d a_i S\big(e_i^{\otimes n}\big) + \sum_{i\neq j} b_{i,j} S\big( e_i^{\otimes (n-1)} \otimes e_j \big) + \sum_{i\neq j,k} c_{i,j,k} S\big(e_i^{\otimes (n-2)}\otimes e_j \otimes e_k\big) + \cdots$$
$P(x_1,x_2,\cdots, x_d)= P(\mathbf{x})$ plays the role of the $\alpha$ in (\ref{Polar}) or (\ref{alpha}).)
This problem probably admits different solutions: (already the case of the decomposition of quadratic form as a sum of squares. The parallelograms identity is in fact an equality of two sums of squares!)


*

*Apply (\ref{Polar}) for $\alpha: \mathbb{R}^n \to \mathbb{R},\ (y_1,\cdots, y_n) \mapsto \prod_{j=1}^n y_j$ yields
$$ y_1 \cdots y_n= \frac{1}{n!}\sum_{1\leq i_1 < i_2 < \cdots < i_k \leq n} (-1)^{n-k} \big(y_{i_1} + y_{i_2} + \cdots + y_{i_k}\big)^n \tag{Polar2}\label{Polar2}$$
and successively replacing $y_1 \cdots y_n$ by the monomials $x_i^n,\ x_i^{n-1} x_j,\ x_i^{n-2} x_j x_k$ etc. of (\ref{Poly}) will yield (\ref{nPower}). This is what seems to be done here but this other answer looks much more interesting.

*Instead of doing it for each monomial, one can try to treat the problem one variable $x_i$ after another: assume that one of the $a_i$ is non-zero (otherwise, jump to the other cases which have to be treated anyway). Let's assume it is $a_1$, then
$$P(x_1,x_2,\cdots, x_d)= a_1 x_1^n +  x_1^{n-1} B(x_2,\cdots, x_d) + x_1^{n-2}  C(x_2,\cdots, x_d) + \cdots \tag{a}\label{Fctze}$$
where $B$ is a polynomial of order 1, $C$ of order 2 etc. on the $n-1$ other variables.
$$ \ref{Fctze} = a_1 \left(x_1 + \frac{B(x_2,\cdots, x_d)}{na_1}\right)^n - x_1^{n-2}\left( C(x_2,\cdots, x_d) - {n\choose 2} \Big(\frac{B(x_2,\cdots, x_d)}{na_1}\Big)^2 \right) + \cdots $$
The second term is of the form $ x_1^{n-2}\ \tilde{C}(x_2,\cdots, x_d)$ with $\tilde{C}$ quadratic. Use a decomposition of it as sum of squares $\tilde{C}= \sum c_p l_p(x_2,\cdots, x_d)^2$. Inspired by (\ref{Polar2}), one guesses that
$$x_1^{n-2} l_p^2 = \big((n-2)x + 2 l_p\big)^n - 2\big((n-2)x + l_p \big)^n -(n-2) \big((n-3)x + 2 l_p \big)^n + \cdots $$
not sure I can get an explicit formula...
