Polynomial and super-symmetric tensor A quadratic function uniquely determines a symmetric matrix. Ok that’s easy. Now a homogeneous polynomial function $f(x)$ also uniquely determines a super-symmetric tensor.
My question is how do I have to build this tensor? For example I have:
$$f(s,t) = s^4 + 7 s^3 t + 3s t^3 +9t^4$$
How to make a super-symmetric tensor out of that?
 A: The way I know is the following:
Let $f(x_1,\ldots,x_n)$ be any homogeneous polynomial of the variables $x_1,\ldots,x_n$ and of degree $m \in \mathbb N$. For simplicity I will assume that the coefficients of $f$ are real. Each monomial in $f$ is of the form $$c_{i_1,\ldots,i_m}x_{i_1}\cdot \ldots \cdot x_{i_m} \quad \text{ with } \quad c_{i_1,\ldots,i_m} \in \mathbb R.$$
Let us recall that a squared tensor $T \in \mathbb R^{n \times \ldots \times n}$ is supersymmetric if and only if $T_{i_1,\ldots,i_m} =T_{\sigma(i_1),\ldots,\sigma(i_m)}$ for every permutation $\sigma \in \mathfrak{G}_n$ of $m$ elements (see the symmetric group). Furthermore we may consider the function
$$f_T: \mathbb R^n \to \mathbb R: x \mapsto \sum_{i_1 = 1,\ldots,i_m=1}^n T_{i_1, \ldots i_m}x_{i_1}\cdot \ldots \cdot x_{i_n}.$$
It is now clear that $f_T=f$ if $$T_{i_1,\ldots,i_m} = \frac{c_{i_1,\ldots,i_m}}{s(i_1,\ldots,i_m)},\quad \text{with } \quad s(i_1,\ldots, i_m) = \left|\bigcup_{\sigma \in \mathfrak G_n} \sigma(i_1,\ldots,i_m)\right|$$
for every $i_1,\ldots, i_m= 1,\ldots, n$, where $|S|$ denotes the cardinality of the set $S$ (which is also the number of different element in $S$).
For your example consider the tensor $T \in \mathbb{R}^{2 \times 2 \times 2 \times 2}$ where $$T_{i_1,i_2,i_3,i_4} = \left\{\begin{array}{l l} 1 & \text{ if } i_1 = \ldots = i_4 = 1\\ 9 & \text{ if } i_1 = \ldots = i_4 = 2\\ 7/4& \text{ if } |\{k:i_k = 1\}|  = 3  \text{ and }|\{k:i_k = 2\}|  = 1 \\ 3/4 & \text{ if } |\{k:i_k = 1\}|  = 1  \text{ and }|\{k:i_k = 2\}|  = 3 \\ 0 & \text{else. } \end{array}\right.$$
