that the product appears simpler when we multiply in an order which, when reading from left to right, goes *against* the direction of an arrow. In this case, we mean that the dimension vector for $f$ is supported on the subquiver $Q^1$ consisting only of the vertex $1$, while the dimension vector for $g$ is supported on the subquiver $Q^2 = 2 \leftarrow 3$, and these two subquivers have the property that $Q^1$ sits at the head of an arrow from $Q^2$.

\[ex:coha.mult.again\] Again we let $Q$ be the quiver $1 \leftarrow 2 \leftarrow 3$. This time, take $\omega_{2,1} \in {\mathscr{H}}_{(0,1,0)}$ and $\omega_{3,1} \in {\mathscr{H}}_{(0,0,1)}$ to see that $$\begin{aligned}
        \omega_{2,1} * \omega_{3,1} &= \omega_{2,1}\omega_{3,1} \in {\mathscr{H}}_{(0,1,1)} \\
        \omega_{3,1} * \omega_{2,1} &= \omega_{2,1}\omega_{3,1} (\omega_{2,1} - \omega_{3,1}) \in {\mathscr{H}}_{(0,1,1)}
    \end{aligned}$$ where we again observe that going “against” the direction of an arrow (this time the arrow $2 \leftarrow 3$) makes the multiplication simpler. Now, consider $\omega_{3,1}$ not as an element of ${\mathscr{H}}_{(0,0,1)}$, but instead as an element of ${\mathscr{H}}_{(0,1,1)}$. This results in the products $$\begin{aligned}
        \omega_{2,1} * \omega_{3,1} &= -\omega_{3,1} \in {\mathscr{H}}_{(0,2,1)} \\
        \omega_{3,1} * \omega_{2,1} &= (\omega_{2,1}+\omega_{2,2})\,\omega_{3,1} - \omega_{3,1}^2 \in {\mathscr{H}}_{(0,2,1)},
    \end{aligned}$$ illustrating the dependence on the dimension vector grading.

Geometric 