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A stability degree bounds the first row of every padding tail

Statement

Read missing partition parts as 0 when comparing row lengths.

Let k be a field of characteristic 0, let V be an FI-module over k (The category FI and FI-modules over a commutative ring), and put s:=stab-deg⁡(V)<∞ (Injectivity, surjectivity and stability degree). Then for every n≥0, every irreducible constituent V(λ)n of the kSn-module Vn satisfies

λ1≤s.

Equivalently, since the padded partition is λ[n]=(n−∣λ∣,λ1,λ2,… ), the second row of the Specht diagram λ[n] has length at most s; the first row of the padding tail is the second row of the full diagram (First-row padding of a partition, Weight of an FI-module over a field of characteristic zero). No choice principle is used.

Facts & Assumptions

Given: A field k of characteristic 0, an FI-module V over k with s=stab-deg⁡(V)<∞, and for m≥0 the sub-FI-module FmV:=span⁡V(V≤m), with F−1V:=0, the graded quotient Gm:=FmV/Fm−1V and its degree-m part Hm:=(Gm)m; for a partition μ⊢m, Vμ is the irreducible kSm-module with label μ (Specht modules classify the irreducibles over every field of characteristic zero, Weight of an FI-module over a field of characteristic zero, Quotient module M/N with scalar multiplication on additive cosets).

[F1]

span⁡V(V≤m) is the smallest sub-FI-module containing all values of V in degrees ≤m; it is generated in degree ≤m, the inclusions FmV⊆Fm+1V hold, and V=⋃m≥0FmV because every element of VS is the bijection transport of an element of V[∣S∣] and hence lies in (F∣S∣V)S (Span, finite generation and generation degree of an FI-module).

[F2]

Every quotient U↠W of FI-modules satisfies surj-deg⁡(W)≤surj-deg⁡(U) (Bounded generation degree bounds surjectivity degree and weight); generation in degree ≤d passes to quotients, and for such a quotient the evaluation map ⨁i≤dM(Wi)↠W is surjective (Generation in bounded degree presents an FI-module as a quotient of free FI-modules, The free FI-module M(W) generated by an Sm-module).

[F3]

For μ⊢m and 0≤a≤m, the coinvariants (Vμ)Sm−a vanish if and only if m−a>μ1, i.e. μ1<m−a (Coinvariants detect vanishing of subquotients of bounded weight, and padded irreducibles stabilise, The coinvariant functors Φa and the transition maps T).

[F4]

For a partition μ⊢m and n≥m, the constituents of M(Vμ)n≅Ind⁡Sm×Sn−mSn(Vμ⊠k) are the Vν with ν⊢n obtained from μ by adding one box in each of n−m different columns, so ν1≥ν2 and ν2≤μ1; and if ν=λ[n] then ν2=λ1 (Branching rule for Specht modules: adding or removing boxes in distinct columns, The free FI-module M(W) generated by an Sm-module, First-row padding of a partition).

[F5]

Φa(V)n=(Va⊔n)Sn, and T ⁣:Φa(V)n→Φa(V)n+1 is the map induced by an injection a⊔n→a⊔n+1 fixing a; stab-deg⁡(V)=max⁡(inj-deg⁡(V),surj-deg⁡(V)), so surj-deg⁡(V)≤s (The coinvariant functors Φa and the transition maps T, Injectivity, surjectivity and stability degree).

[F6]

Over the characteristic-zero field k, every finite-dimensional kSm-module is a direct sum of irreducibles Vμ (Maschke's theorem for finite groups over fields whose characteristic does not divide ∣G∣, Specht modules classify the irreducibles over every field of characteristic zero), and M is additive in its argument, so M(⨁iNi)≅⨁iM(Ni) (The free FI-module M(W) generated by an Sm-module).

Proof

technique · direct
1.1F1given

The FmV are sub-FI-modules with FmV⊆Fm+1V, V=⋃mFmV, and each FmV is generated in degree ≤m.

1.2F1F6givenalgebra

If every constituent of every (Gm)n has padding tail first row ≤s, then so does every constituent of every Vn: let V(λ)n≅P/Q be a constituent of Vn and lift a k-basis of the finite-dimensional module P/Q to elements of P and take the span of their finite Sn-orbits, giving a finite-dimensional kSn-submodule P0⊆P with P0+Q=P; then P0/(P0∩Q)≅P/Q, and P0⊆(FMV)n for some M because a finite basis of P0 lies in a common filtration term. Intersecting the filtration with P0 gives finite-dimensional successive factors embedding in (Gm)n. Maschke splits this finite filtration, so every simple constituent of P0 occurs in one of those factors; hence V(λ)n is a constituent of some (Gm)n.

1.3given

Suppose m≤s. Then every constituent Vμ of Hm has μ1≤m≤s because μ⊢m.

1.4F2F3F5given

Suppose m≥s+1 and set V‾:=V/Fm−1V and a0:=m−s−1≥0. As a quotient of V, the FI-module V‾ has surj-deg⁡(V‾)≤s by [F2], so T ⁣:Φa0(V‾)s→Φa0(V‾)s+1 is surjective. Its domain is (V‾m−1)Ss=0 because V‾m−1=0, and its target is (V‾m)Ss+1; also V‾m=Vm/span⁡V(V<m)m≅Hm by the second isomorphism theorem. Hence (Hm)Ss+1=0, and for every constituent Vμ of Hm the coinvariants (Vμ)Ss+1 are a subquotient of (Hm)Ss+1, hence vanish; by [F3] applied with a=m−(s+1), so that m−a=s+1, this says s+1>μ1, i.e. μ1≤s.

2.1F1F2step 1.1

The quotient Gm is generated in degree ≤m and has (Gm)i=0 for i<m, so by [F2] the evaluation map M(Hm)↠Gm is surjective. Hence every constituent of (Gm)n is a constituent of M(Hm)n.

2.2F1F4F6step 1.4

Every constituent of M(Hm)n has padding tail first row ≤s: let V(λ)n≅P/Q be a constituent of M(Hm)n and choose x∈P∖Q; the Sn-orbit span X of x is finite-dimensional, and X/(X∩Q) is a nonzero submodule of the simple V(λ)n, hence isomorphic to it, so V(λ)n is a constituent of X. Since X lies in the direct sum ⨁T⊆[n],∣T∣=mHm(T) over finitely many T, after transporting by bijections [m]→T for these finitely many subsets, the projections of X to the summands Hm(T) generate a finite-dimensional kSm-submodule N⊆Hm with X⊆M(N)n; so V(λ)n is a constituent of M(N)n. By [F6] write N≅⨁μVμ⊕cμ (finite sum over partitions μ⊢m with cμ≠0); every such Vμ is a constituent of Hm, so μ1≤s by steps 1.3 and 1.4. By [F6] M(N)≅⨁μM(Vμ)⊕cμ, and by [F4] every constituent Vν of M(Vμ)n satisfies ν2≤μ1≤s; a constituent of a finite direct sum is a constituent of one summand, so for the constituent V(λ)n≅Vλ[n] of M(N)n we get λ1=(λ[n])2≤s.

3.1step 1.2step 1.3step 1.4step 2.1step 2.2∎

Steps 2.1 and 2.2 show that every constituent of every (Gm)n has padding tail first row at most s (for m≤s via step 1.3, for m≥s+1 via step 1.4), and step 1.2 transfers this bound to every constituent V(λ)n of Vn; thus λ1≤s always. Since λ[n]=(n−∣λ∣,λ1,λ2,… ), the number λ1 is the length of the second row of the Specht diagram λ[n]. No choice was used.

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