are not imposing any IBP or EOM relations in this counting. The $q$-Pochhammer symbol gives us a clear interpretation of the series coefficients in terms of partitions, which we return to at the end of this subsection.

Let us show how the above Hilbert series is obtained using the framework of section \[sec:framework\]. For clarity, we consider the $N = 1$ case; generalization to arbitrary $N$ is straightforward. For fixed $r$, the module is $R^{S_r} = \mathbb{R}[P_1,\dots,P_r]$, where the $P_n$ carry weight $t^n$. The number of degree $k$ polynomials in $R^{S_r}$ is number of points in the set $\{(k_1,\dots,k_r)~|~k_1 + 2k_2 + \dots + rk_r = k\}$ so that the Hilbert series is $$H_{r,\text{free}}(R^{S_r},t) = \sum_{k=0}^{\infty}t^k\sum_{k_1+2k_2+\dots+rk_r = k} = \sum_{k_1=0}^{\infty}\dots \sum_{k_r=0}^{\infty} t^{k_1 + 2k_2 + \dots +rk_r} = \frac{1}{\prod_{n=1}^{r}(1-t^n)}.
\label{eqn:Hil_symm_poly}$$ The above result is made transparent by recognizing that all independent monomials in $R^{S_r}$ are obtained in the expansion of $(1+P_1 +P_1^2+\dots)(1+P_2+P_2^2+\dots)\dots(1+P_r+P_r^2+\dots)$; substituting $P_n$ with its weight $t^n$ and geometrically summing produces the above result. The full Hilbert series is $$H_{1,\text{free}}(t,u) = \sum_{r=0}^{\infty} \frac{u^r}{\prod_{n=1}^{r}(1-t^n)} = \frac{1}{(u;t)_{\infty}},$$ reproducing eq.  for $N=1$.

Let us now give an interpretation of the coefficients in the expansion of the Hilbert series, $H_{N,\text{free}} = \sum_k \sum_{\br} 