Since $f_1 \in {\mathscr{H}}_{\gamma_1}$ (respectively $f_2\in {\mathscr{H}}_{\gamma_2}$) it is a polynomial separately symmetric in $n$ families of indeterminates, one family for each $i\in Q_0$, each with $\gamma_1(i)$ (resp. $\gamma_2(i)$) variables. Hence, it makes sense to let $f_1(S_\bullet)$ denote the specialization $f_1(\omega_{i,j})$ for all $i\in Q_0$ and $j\in S_i$ (resp. $f_2(\overline{S_\bullet})$ denotes the specialization $f_2(\omega_{i,j})$ for all $i\in Q_0$ and $j\in \overline{S_i}$). Finally, the products in the numerator and denominator should be interpreted according to the convention $$\label{eqn:prod.set.subscripts}
    \left(\omega_{x,U} - \omega_{y,V}\right) := \prod_{u\in U} \prod_{v\in V} \left(\omega_{x,u} - \omega_{y,v}\right).$$ We comment that if $U$ or $V$ is empty, then the value of the product is $1$. From the supersymmetry of $f_1$ and $f_2$, and the fact that we sum over all possible choices for $(S_1,\ldots,S_n)$, we see that the result of will indeed have the supersymmetry required to be an element of ${\mathscr{H}}_{\gamma_1+\gamma_2}$.

\[ex:coha.mult\] Let $Q$ be the equioriented $A_3$ quiver $1 \leftarrow 2 \leftarrow 3$ as in Examples \[ex:Reineke.order.A3\] and \[ex:Orbit.A3\]. Consider $f(\omega_{1,1},\omega_{1,2}) \in {\mathscr{H}}_{(2,0,0)}$ and $g(\omega_{2,1},\omega_{3,1}) \in {\mathscr{H}}_{(0,1,1)}$. We have that $$\begin{aligned}
    (f*g)(\omega_{1,1},\omega_{1,2},\omega_{2,1},\omega_{3,1}) 
        &= f\left(\omega _{1,1},\omega _{1,2}\right) g\left(\omega _{2,1},\omega _{3,1}\right)\\
    (g*f)(\omega_{1,1},\omega_{1,2},\omega_{2,1},\omega_{3,1}) 
        &=  f\left(\omega _{1,1},\omega _{1,2}\right) g\left(\omega _{2,1},\omega _{3,1}\right) \left(\omega _{1,1}-\omega _{2,1}\right) \left(\omega _{1,2}-\omega _{2,1}\right).
    \end{aligned}$$ From this we observe 