Murnaghan Stability via FI-Modules — Examples

Rank four from characters. By [F1], the five $S_4$ Schur functions are $s_{(4)}=h_4$, $s_{(3,1)}=h_3{h}_1-h_4$, $s_{(2,2)}=h_2^2-h_3{h}_1$, $s_{(2,1,1)}=h_2{h}_1^2-h_2^2-h_3{h}_1+h_4$, and $s_{(1,1,1,1)}=h_1^4-3h_2{h}_1^2+h_2^2+2h_3{h}_1-h_4$. Substituting the complete-function expansions and multiplying each $p_{\rho }$ coefficient by $z_{\rho }$ gives the character rows $(1,1,1,1,1)$, $(3,1,-1,0,-1)$, $(2,0,2,-1,0)$, $(3,-1,-1,0,1)$, $(1,-1,1,1,-1)$ on the cycle types $(1^4),(2,1,1),(2,2),(3,1),(4)$ respectively; here $z_{\rho }=(24,4,8,3,4)$. The squared $(2,2)$ character row is $(4,0,4,1,0)$. The triple-character sums are $g^{(4)}=4\mathrm{/}24+4\mathrm{/}8+1\mathrm{/}3=1$, $g^{(1,1,1,1)}=4\mathrm{/}24+4\mathrm{/}8+1\mathrm{/}3=1$, $g^{(2,2)}=8\mathrm{/}24+8\mathrm{/}8-1\mathrm{/}3=1$, and $g^{(3,1)}=g^{(2,1,1)}=12\mathrm{/}24-4\mathrm{/}8=0$. Hence [F2] gives $s_{(2,2)}\ast s_{(2,2)}=s_{(4)}+s_{(1,1,1,1)}+s_{(2,2)}$. The dimension check is $1+1+2=4=2^2$.

All six products from local straightening. Apply $U_n$ to the ten-term polynomial of [F3]. The five tails $\mathrm{\varnothing },(1),(2),(1,1),(1,1,1)$ are already admissible for every $n\ge 4$. For the remaining five tails, sorting the $q_j$ of [F4] gives the following values at ranks $4,5,6,7$; each becomes its ordinary padded Schur function once its threshold $\mathrm{\mid }\gamma \mathrm{\mid }+{\gamma }_1$ is reached: $\begin{array}{ccccc}\gamma & 4& 5& 6& 7\\ (2,1)& 0& s_{(2,2,1)}& s_{(3,2,1)}& s_{(4,2,1)}\\ (2,2)& 0& 0& s_{(2,2,2)}& s_{(3,2,2)}\\ (3)& -s_{(2,2)}& 0& s_{(3,3)}& s_{(4,3)}\\ (3,1)& -s_{(2,1,1)}& -s_{(2,2,1)}& 0& s_{(3,3,1)}\\ (4)& -s_{(3,1)}& -s_{(3,2)}& -s_{(3,3)}& 0\end{array}$. Multiplying by the coefficients in [F3] and collecting terms yields precisely the four displayed products at ranks $4$ through $7$. At ranks $8,9$ every tail is admissible, and padding the same polynomial yields precisely the two displayed products, including zeros for every unlisted Schur label. This verifies the six products without assuming the source table.

The reduced coefficients and stable ranks. By [F3], the ten coefficients of $R_{(2),(2)}$ are exactly the listed reduced coefficients, and all others vanish. For $n\ge 8$, [F4] sends each padded label to its rank-$(n+1)$ padding without changing its coefficient. Thus the rank-eight product maps to the rank-nine product term by term, and [F3] proves stability for all later ranks, with exact index $8=2+2+2+2$.

Failure at rank seven. The rank-seven expansion has the nine labels $(7),(4,1,1,1),(5,2),(6,1),(5,1,1),(4,2,1),(3,2,2),(4,3),(3,3,1)$, with coefficients $1,1,2,1,1,2,1,1,1$. Increasing their first parts by one gives exactly the rank-eight expansion minus $s_{(4,4)}$. That last term has coefficient one and is outside the image of $V$, since every $V(s_{\delta })$ has strictly larger first part than second part. Hence the rank-eight product is not $V$ of the rank-seven product.

The individual coefficient. The computed rank-four and rank-five products have coefficient one at their $(n-2,2)$ labels; ranks six and seven have coefficient two, and every $n\ge 8$ has the reduced coefficient two by step 2.1. Thus this coefficient first becomes constant at rank six, while the whole product first becomes $V$-stable at rank eight. All computations and comparisons are finite, and no choice principle is used.

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.