Bijective association between the linear space of symmetric tensors and linear space of homogeneous polynomials example edit:
Given a symmetric tensor G there is a bijective mapping between the $d$ indexes and a vector of $K$ numbers representing the time each variable $x_t$ appears in the associated monomial. For example, imagine $G_{112}$ being a 3 dimensional tensor with every index spanning from 1 to 3 (so $G_{333}$ is the last element)
The element is associated with $G_{112}x_1^2x_2^1x_3^0 = G_{112}x^{f([1,1,2])}$ where $f([1,1,2]) = [2,1,0]$.
More precisely
\begin{equation}
p(x) = \sum G_ix^{f(i)}
\end{equation}
So i want to "see" this relation using an example so i took two 2-dimensional symmetric tensors like \begin{pmatrix}
1 & 0 \\
0 & 1
\end{pmatrix}
which is associated with the homogeneous polynomial
\begin{equation}
p(x_1,x_2) =x^2_1+x^2_2
\end{equation}
and
\begin{pmatrix}
0 & 1 \\
1 & 0
\end{pmatrix}
which is associated with the homogeneous polynomial
\begin{equation}
q(x_1,x_2) = 2x_1x_2
\end{equation}
Now if i do the product
\begin{equation}
p(x_1,x_2)q(x_1,x_2) = 2x_1^3x_2 + 2x_1x_2^3
\end{equation}
i should obtain a homogeneous polynomial which is associated to a symmetric tensor, and that symmetric tensor is
\begin{equation}
\begin{pmatrix}
0 & \frac{1}{2} \\
\frac{1}{2} & 0
\end{pmatrix}
\begin{pmatrix}
\frac{1}{2} & 0 \\
0 & \frac{1}{2}
\end{pmatrix}
\begin{pmatrix}
\frac{1}{2} & 0 \\
0 & \frac{1}{2}
\end{pmatrix}
\begin{pmatrix}
0 & \frac{1}{2} \\
\frac{1}{2} & 0
\end{pmatrix}
\end{equation}
But i can also calculate the tensor product of the two original tensors which is 4-dimensional and is
\begin{equation}
\begin{pmatrix}
0 & 1 \\
1 & 0
\end{pmatrix}
\begin{pmatrix}
0 & 0 \\
0 & 0
\end{pmatrix}
\begin{pmatrix}
0 & 0 \\
0 & 0
\end{pmatrix}
\begin{pmatrix}
0 & 1 \\
1 & 0
\end{pmatrix}
\end{equation}
They should be the same... What am i doing wrong? Can you give an example of this link between tensors and polynomials?

This is because the tensor product of symmetric tensors is not necessarily a symmetric tensor. In your example, the rank-$4$ tensor
$$
T = \begin{pmatrix}
\begin{pmatrix}
0 & 1 \\
1 & 0
\end{pmatrix} &
\begin{pmatrix}
0 & 0 \\
0 & 0
\end{pmatrix}\\
\begin{pmatrix}
0 & 0 \\
0 & 0
\end{pmatrix} &
\begin{pmatrix}
0 & 1 \\
1 & 0
\end{pmatrix}\\ 
\end{pmatrix}
$$
is not symmetric. For example, the entries $T_{1112} = 1$ and $T_{1211} = 0$ are not equal.
There's another indication that simply taking the tensor product doesn't work, i.e. tensor products of symmetric tensors are not commutative, but multiplying polynomials certainly is.
I'm not familiar with this mapping between homogeneous polynomials and symmetric tensors that you have described, but I think what you want is probably the symmetric product, which means that after taking the tensor product you need to average all entries with indices in the same orbit w.r.t. permutation. Note that the symmetric product is also commutative, which is a good sign. Again taking the $1112$ set of indices as an example, we have $T_{1112} = 1$, $T_{1121} = 1$, $T_{1211} = 0$, $T_{2111} = 0$, so the symmetrized version should have $T_{(1112)} = \frac{1}{2}$. You can check that this gives you the same matrix as the one obtained from the product of the polynomials.
