A global product threshold is not a per-coefficient sharpness claim

Ranks six, seven and eight. In the displayed products of [F3]: $s_{4,2}\ast s_{4,2}$ contains $2s_{4,2}$ (rank $6$), $s_{5,2}\ast s_{5,2}$ contains $2s_{5,2}$ (rank $7$), and $s_{6,2}\ast s_{6,2}$ contains $2s_{6,2}$ (rank $8$). Hence the coefficient $g_{(4,2),(4,2)}^{(4,2)}=g_{(5,2),(5,2)}^{(5,2)}=g_{(6,2),(6,2)}^{(6,2)}=2$ equals the reduced value ${\bar{g}}_{(2),(2)}^{(2)}=2$ at ranks $6,7,8$.

Rank five is different. The rank-five product $s_{3,2}\ast s_{3,2}$ contains $s_{3,2}$ with coefficient $1$, not $2$, so the coefficient $g_{(3,2),(3,2)}^{(3,2)}=1\mathrm{\ne }{\bar{g}}_{(2),(2)}^{(2)}$; together with step 1.1 this shows that the least rank from which the coefficient equals the stable value for all later ranks is $6$. For $n\ge 6$ this is indeed stable: ranks $6,7,8$ are displayed in step 1.1 and ranks $n\ge 8$ follow from [F1] with the padded label $(n-2,2)=\gamma [n]$.

Comparison with the global threshold. By [F2] the whole product is $V$-stable exactly from rank $8$ onward, and it is not $V$-stable at rank $7$; the single coefficient of steps 1.1 and 2.1 has already reached its stable value $2$ at rank $6$ and stays there. Since $6<8$, the global product threshold is strictly larger than the stabilization rank of this coefficient, so no statement of the form "each coefficient first becomes constant at the global threshold" can be true. This is a finite boundary witness; it uses the displayed products, the exact threshold theorem and the reduced value $2$, all established above.