Weight of an FI-module over a field of characteristic zero

Let $k$ be a field of characteristic $0$ and let $V$ be an FI-module over $k$ (The category FI and FI-modules over a commutative ring). For a partition $\lambda$ and $n\ge \mathrm{\mid }\lambda \mathrm{\mid }+{\lambda }_1$ put $\lambda [n]:=(n-\mathrm{\mid }\lambda \mathrm{\mid },{\lambda }_1,{\lambda }_2,\dots )$, the first-row padding (First-row padding of a partition), with $\mathrm{\mid }\mathrm{\varnothing }\mathrm{\mid }={\mathrm{\varnothing }}_1=0$ and $\mathrm{\varnothing }[0]=\mathrm{\varnothing }$, and let

$V(\lambda {)}_n:=k{\otimes }_{\mathbb{Q}}{S}_{\mathbb{Q}}^{\lambda [n]}$

be the scalar extension of the rational Specht model of $\lambda [n]$ (Complex Specht modules are irreducible, Specht modules classify the irreducibles over every field of characteristic zero). By the classification lemma, $V(\lambda {)}_n$ is the irreducible $k{S}_n$-module with label $\lambda [n]$, and every irreducible $k{S}_n$-module is isomorphic to exactly one $V(\lambda {)}_n$: indeed, every partition $\mu =({\mu }_1,{\mu }_2,\dots )$ of $n$ is uniquely $\lambda [n]$ for the partition $\lambda =({\mu }_2,{\mu }_3,\dots )$ obtained by deleting the first row, and $n\ge \mathrm{\mid }\lambda \mathrm{\mid }+{\lambda }_1$ holds precisely because ${\mu }_1\ge {\mu }_2$; consequently the labels $\lambda [n]$ run over the partitions of $n$ without repetition. At $n=0$ the only label and its tail are both $\mathrm{\varnothing }$, and $V(\mathrm{\varnothing }{)}_0=k$ with the trivial $S_0$-action.

Say that $V(\lambda {)}_n$ occurs as a constituent of $V_n$ when there are $k{S}_n$-submodules $U\subseteq W\subseteq V_n$ with $W\mathrm{/}U\cong V(\lambda {)}_n$ (Submodule of a module, Quotient module $M\mathrm{/}N$ with scalar multiplication on additive cosets); this is a condition on subquotients and requires no decomposition or finite-dimensionality of $V_n$. The weight of $V$ is

$\mathrm{weight}(V):=\sup \{\mathrm{\mid }\lambda \mathrm{\mid }:V(\lambda {)}_n\text{ occurs as a constituent of }{V}_n\text{ for some }n\ge \mathrm{\mid }\lambda \mathrm{\mid }+{\lambda }_1\},$

the supremum over all pairs $(\lambda ,n)$ as displayed. The set of integers displayed is a set of nonnegative integers, so the supremum is well defined in $\mathbb{N}\cup \{\mathrm{\infty }\}$; it is $0$ for $V=0$, and $\mathrm{weight}(V)=\mathrm{\infty }$ exactly when the displayed set is unbounded. No finiteness of $\mathrm{weight}(V)$ is asserted.

If $W$ is a subquotient of $V$, meaning that $W\cong U^{\mathrm{\prime }}\mathrm{/}U$ for sub-FI-modules $U\subseteq U^{\mathrm{\prime }}\subseteq V$ computed pointwise, then $\mathrm{weight}(W)\le \mathrm{weight}(V)$: a constituent of $W_n$ is a subquotient of the subquotient $U_n^{\mathrm{\prime }}\mathrm{/}{U}_n$ of $V_n$, hence a subquotient of $V_n$, so it contributes the same label $\mathrm{\mid }\lambda \mathrm{\mid }$ to the supremum for $V$. No choice principle is used: the labels $\lambda [n]$, constituents and the supremum are set-theoretic data, and no bases or decompositions are chosen.