Spencer support signs, wall reflections, and nonzero composition scalars

Suppose $\mathrm{\ell }<\mathrm{\infty }$ for the digit and reflection arguments. The digit bounds give ${\sum }_{j<k}{n}_j{s}_j\le s_k-1$ for every available $k\ge 1$: for $k=1$ this is $n_0\le \mathrm{\ell }-1$, and adjoining a digit at place $s_k$ increases the maximum from $s_k-1$ to $(p-1)s_k+s_k-1=s_{k+1}-1$. In characteristic zero there is only the first such bound to check. Therefore the signed sum with positive highest digit is at least $n_r{s}_r-(s_r-1)-1\ge 0$ and at most $N$. Flipping signs changes $N$ by an even integer. For $r=0$, the only support element is $N$, so the same assertions hold.

We prove the quantum-binomial zero test needed below, retaining scalar units. Using [F4] when necessary, adjoin a root $q\mathrm{\ne }0$ of $X^2-\delta X+1$ if necessary and put $Q=q^2$. If $Q\mathrm{\ne }1$, the recurrence gives $[t{]}_{\delta }=q^{1-t}(1-Q^t)\mathrm{/}(1-Q)$, so finite $\mathrm{\ell }$ is the exact order of $Q$. If $Q=1$, the recurrence gives $[t{]}_{\delta }=t{q}^{t-1}$, so finite $\mathrm{\ell }$ equals the field characteristic. The extension does not change which elements of the original field vanish. Expanding by subsets gives ${\prod }_{i=0}^{u-1}(1+Q^{i}X)={\sum }_v{Q}^{v(v-1)\mathrm{/}2}{G}_{u,v}(Q)X^v$. In the first case, the length-$\mathrm{\ell }$ factor is $1-(-X{)}^{\mathrm{\ell }}$ by the distinct roots of $Y^{\mathrm{\ell }}-1$; in the second it has this same value by [F3].

If two sign choices differ at a nonzero digit, let $j$ be their largest differing index. Their signed sums differ by a term of absolute value $2n_j{s}_j\ge 2s_j$, whereas all smaller terms have combined absolute value at most $2(s_j-1)$ by step 1.1. Thus their sums differ. Each of the $h$ nonzero lower digits consequently gives an independent choice, proving the count.

If $\mathrm{\ell }=\mathrm{\infty }$, every quantum factorial is nonzero and [F2] proves the asserted nonvanishing directly. Suppose $\mathrm{\ell }<\mathrm{\infty }$ and write $u=U\mathrm{\ell }+u_0$ and $v=V\mathrm{\ell }+v_0$ with remainders in $[0,\mathrm{\ell }-1]$. Step 1.2 factors its product into $(1-(-X{)}^{\mathrm{\ell }}{)}^U{\prod }_{i=0}^{u_0-1}(1+Q^{i}X)$. Comparing the coefficient of $X^v$, and using [F2], shows $B_{u,v}(\delta )\mathrm{\ne }0$ precisely when $v_0\le u_0$ and $\left(\genfrac{}{}{0px}{}{U}{V}\right)\mathrm{\ne }0$ in the field: the omitted factors are powers of $Q$ and $-1$, and $G_{u_0,v_0}$ is nonzero because every quantum factorial index is less than $\mathrm{\ell }$. In positive characteristic, expand $(1+X{)}^U={\prod }_i(1+X^{p^i}{)}^{u_i}$ using [F3]. The coefficient at $V$ is nonzero exactly when its base-$p$ digits are at most the corresponding digits of $U$, since then it is a product of nonzero binomials with factorial indices less than $p$. In characteristic zero $\left(\genfrac{}{}{0px}{}{U}{V}\right)$ is nonzero whenever $0\le V\le U$. Consequently the quantum binomial is nonzero exactly when every mixed-radix digit of $v$ is at most that of $u$.

For $\mathrm{\ell }<\mathrm{\infty }$ and a fixed split index $k$, the higher contribution $H$ is divisible by $s_k$ and satisfies $H\ge n_r{s}_r-{\sum }_{k\le j<r}{n}_j{s}_j\ge s_k>0$ by the same place-value bound. Step 1.1 gives $\mathrm{\mid }L\mathrm{\mid }<s_k$. Negating exactly the lower signs replaces $H+L-1$ by $H-L-1=2(H-1)-(H+L-1)$, retains the highest positive sign, and hence stays in the support. This proves the reflection assertion, including the case $L=0$.

For the nested sets in the Statement, subtracting the signed sums gives $(a-c)\mathrm{/}2={\sum }_{j\in C\setminus A}{n}_j{s}_j$ and $(a-b)\mathrm{/}2={\sum }_{j\in B\setminus A}{n}_j{s}_j$. These are their actual mixed-radix expansions, because each digit is either $0$ or the original allowed digit and no position is added twice. Since $B\setminus A\subseteq C\setminus A$, the second expansion is digitwise bounded by the first. Step 2.2 proves the nonzero scalar, including repeated labels and zero digits. If $\mathrm{\ell }=\mathrm{\infty }$ then the singleton support gives $B_{0,0}=1$ directly from [F2].