Spencer decomposition numbers over arbitrary fields

We first record the triangular facts used below. Let $I_t$ be the ideal spanned by diagrams of propagation at most $t$, with $I_{-1}=I_{-2}=0$. The action on $W_{n,m}$ kills $I_{m-2}$, because propagation cannot increase. On a nonzero $D_{n,r}$, the ideal $I_r$ acts nontrivially: the cellular rule $x{C}_{u,v}^r={\varphi }_{n,r}(x,u)v$ gives a nonzero operator since the form is nonzero. Thus a composition factor of $W_{n,m}$ cannot have label $r<m$. If $D_{n,m}\mathrm{\ne }0$, the same rank-one rule shows that $I_m$ kills the form radical, while $W_{n,m}$ modulo this radical is $D_{n,m}$. Therefore this label has multiplicity exactly one; every other factor has larger label. If $m=0,\delta =0,n>0$, the form is identically zero, so this label contributes no simple quotient. These are all cases because a link state has nonzero form whenever $m>0$ or $\delta \mathrm{\ne }0$: at nonzero loop use its self-pairing, and at positive $m$ use the two shifted adjacent-cup states whose gluing has no loop.

Write $d(N,m)=1$ when $m\in \mathrm{Supp}(N)$ and $0$ otherwise. We prove the lower bound for the trivial-factor multiplicity $t(N,m)=[W_{N,m}:D_{N,N}]$. Call $v_{N,m}$ embodied if there are a submodule $U\subseteq W_{N,m}$ containing it and a module map $q:U\to D_{N,N}$ with $q(v_{N,m})\mathrm{\ne }0$. This implies $t(N,m)\ge 1$. If the support is a singleton, its only label is $N$ and the assertion is immediate. Otherwise write $N+1={\sum }_{i=0}^r{n}_i{s}_i$ with $r\ge 1$ and let $\mu =n_r{s}_r-{\sum }_{i<r}{n}_i{s}_i-1$. The pair $N,\mu$ is a reflection about the $s_r$-wall below $N$, with $0<N-\mu <2s_r$ and $N+\mu +2=2n_r{s}_r$. Thus [F4] gives a nonzero homomorphism $W_{N,N}\to W_{N,\mu }$ carrying its basis vector to $v_{N,\mu }$. Its image is an actual trivial submodule, so this vector is embodied.

We propagate this property across the cube. Consider any subcube with higher signs fixed and all lower nonzero digits up to index $j$ variable. Let $h$ be its maximal label, with all variable signs positive, and $l$ its minimal label, with all variable signs negative; suppose $v_{N,l}$ is embodied. We show all its labels are embodied by induction on the number of variable digits. If there are none, this is the assumption. Otherwise split at its largest variable index $j$. The minus-half has minimum $l$. The plus-half has minimum $a$, obtained by retaining the positive sign at $j$ and making every smaller variable sign negative. Full-prefix reflections give genuine homomorphisms $f:W_{N,h}\to W_{N,a}$ and $H:W_{N,h}\to W_{N,l}$ by [F4], using the identity map for $f$ if $a=h$. All three labels are comparable cube elements, so [F3]–[F4] give $gf=\lambda H$ with $\lambda \mathrm{\ne }0$, where $g:W_{N,a}\to W_{N,l}$ is the linear candidate map. This makes $g$ a module map on $I=\mathrm{im}f$: for any algebra element and any preimage in $W_{N,h}$, the equality with the genuine map $H$ verifies intertwining, and $f$ is surjective onto $I$. Moreover $f(v_{N,h})$ is a nonzero multiple of $v_{N,a}$ and $g(v_{N,a})$ is a nonzero multiple of $v_{N,l}$, again because their candidate-composition scalars are nonzero. If $U_l,q_l$ embody $v_{N,l}$, the submodule $I\cap g^{-1}(U_l)$ and the functional $q_{l}g$ therefore embody $v_{N,a}$. Apply induction to both smaller halves. The whole support cube starts with its embodied minimum from step 2.1, proving $t(N,m)\ge d(N,m)$ for every field and every support label.

We now prove equality by induction on $n$, the case $n=0$ being the one-dimensional algebra. Suppose the full decomposition formula holds at smaller ranks. For each label $m$ put $t_m=t(n,m)$ and let $u_m$ be the multiplicity of the rank-$(n-1)$ trivial module in the restriction of $W_{n,m}$. By [F2] and induction, $u_m=d(n-1,m-1)+d(n-1,m+1)\mathrm{.}$ For each nonzero simple $D_{n,r}$ let $b_r$ be the multiplicity of that trivial module in its restriction; these are nonnegative integers, and $b_n=1$. For $r<n$, [F2] and induction already give $[W_{n,m}:D_{n,r}]=d(r,m)$. It remains only to determine the $t_m$.

Treat labels $m<n$ in decreasing order, assuming $t_r=d(n,r)$ and $b_r={\beta }_{n,r}$ at all larger labels. If $D_{n,m}\mathrm{\ne }0$, step 1.1 and restriction of a composition series give $u_m=t_m+b_m+{\sum }_{m<r<n}d(r,m)b_r\mathrm{.}$ The carry identity [F5] gives $u_m=d(n,m)+{\beta }_{n,m}+{\sum }_{m<r<n}d(r,m){\beta }_{n,r}\mathrm{.}$ Subtracting and using the descending assumption yields $t_m+b_m=d(n,m)+{\beta }_{n,m}$. If ${\beta }_{n,m}=0$, the lower bound of step 3.1 and $b_m\ge 0$ force $t_m=d(n,m)$ and $b_m=0$. If ${\beta }_{n,m}>0$, [F5] makes $W_{n,m}$ simple. Since $m<n$, its simple label is different from $n$, so $t_m=0$; step 3.1 then forces $d(n,m)=0$, and the same equality gives $b_m={\beta }_{n,m}$. This completes the descending step whenever the diagonal simple exists.

The only missing diagonal is $m=0,\delta =0,n>0$ even. Here $D_{n,0}=0$ by step 1.1, so restriction of a composition series gives $u_0=t_0+{\sum }_{0<r<n}d(r,0)b_r$ with no $b_0$ term. By [F5], ${\beta }_{n,0}=0$, and all larger $b_r$ already equal their carry values. The carry identity consequently gives $t_0=d(n,0)$ directly. Thus the exceptional zero-form standard is classified too, and no nonexistent simple has been inserted. The case $n=1$ is covered by the initial top label and needs no descending step.

We have proved $[W_{n,m}:D_{n,n}]=d(n,m)$ at rank $n$, completing the induction by steps 4.1–6.1. Finally, for any nonzero $D_{n,r}$, exact truncation [F2] identifies its desired multiplicity with $[W_{r,m}:D_{r,r}]=d(r,m)$, which is precisely the formula in the Statement. All arguments were over the given field or polynomial identities valid after its extension, and the zero-loop, zero-defect exception was explicitly handled. Hence the formula has every asserted field, characteristic and quantum-characteristic branch.