Difference-one spanning and finite degree spans for cyclic weak modules

Fix $d$. Filter its mode words by total state weight $p\ge 0$, then length $\mathrm{\ell }\ge 0$, in lexicographic order. All reductions below preserve $d$. A homogeneous state $v$ is a linear combination of same-weight representatives and homogeneous $a_{(-2)}b$. Replacing $v_{(n)}$ by the iterate expansion of $(a_{(-2)}b{)}_{(n)}$ replaces its weight $\mathrm{w}\mathrm{t}(a)+\mathrm{w}\mathrm{t}(b)+1$ by the smaller total weight $\mathrm{w}\mathrm{t}(a)+\mathrm{w}\mathrm{t}(b)$. The expansion is finite on the fixed tail vector, by the rightmost-mode truncations in the iterate identity. Thus these replacements lower $p$, irrespective of length. A weight-zero representative is a scalar vacuum; its mode is either zero or the identity and so shortens the word without changing $p,d$. Induction on $p$, and within a fixed $p$ on length, therefore reduces the problem to words in $A$. When $p=0$ only vacuum words occur, reducing to the empty word or zero.

Interchanging adjacent modes uses $[a_{(m)},b_{(n)}]={\sum }_{t\ge 0}\left(\genfrac{}{}{0px}{}{m}{t}\right)(a_{(t)}b{)}_{(m+n-t)}$. Every nonzero commutator state has weight $\mathrm{w}\mathrm{t}(a)+\mathrm{w}\mathrm{t}(b)-t-1$, strictly smaller than the replaced total weight; these errors are already treated by induction on $p$. The commutator sum is finite by state truncation. Bubble sorting the unchanged-length terms therefore reduces any $A$-word modulo lower $p$ to nondecreasing mode indices. A sorted word with last index at least $N$ kills $w$. This cutoff is obtained before any claim of strict indices, and uses only truncation on $w$, not an assumed lower conformal-weight bound.

At fixed $p,\mathrm{\ell },d$, there are only finitely many nondecreasing index tuples with last index less than $N$ that can occur in $A$-words of total state weight $p$. Indeed their sum is $p-\mathrm{\ell }-d$, so each index is at most $N-1$ and at least $p-\mathrm{\ell }-d-(\mathrm{\ell }-1)(N-1)$. This supplies a finite set of tuples with its usual lexicographic order. We may therefore induct upwards in that finite order to remove equal indices. Fix an adjacent repetition $a_{(i)}{b}_{(i)}$ and write its tail as $u$. The normal-product iterate formula is $(a_{(-1)}b{)}_{(2i+1)}u={\sum }_{t\ge 0}(a_{(-1-t)}{b}_{(2i+1+t)}+b_{(2i-t)}{a}_{(t)})u$. If $i<0$, the first sum has exactly one term with both indices $i$, at $t=-i-1$; if $i\ge 0$, the second has exactly one, at $t=i$, with reversed order $b_{(i)}{a}_{(i)}$. In the latter case interchange that pair using step 2.1; its error has lower $p$. Every remaining two-mode term has unequal indices summing to $2i$, hence one less than $i$. Sorting the resulting full word by step 2.1 gives a lexicographically smaller tuple, or zero by its last-index cutoff. The collapsed mode of $a_{(-1)}b$ has the same $p$ and degree as the pair but shorter length, so is already treated by the length induction; replacing its state by representatives if needed is covered by step 1.1. Thus the finite tuple induction removes this repetition. All iterate sums are finite on $u$, and no uniform tail truncation bound is presumed. At fixed filtration the reductions shorten length or lower the finite tuple order; terms at lower filtration were already fully reduced by the outer induction and may have different lengths. All final strings retain degree $d$.

The previous inductions apply to every finite $p,\mathrm{\ell }$, so they span the actual $H_d(w)$, which allows unbounded filtration and length. We now bound the resulting spanning family independently of that filtration. Every representative has weight at least one. A strict length-$\mathrm{\ell }$ tuple with indices less than $N$ satisfies $i_j\le N-1-(\mathrm{\ell }-j)$, so $d={\sum }_j(\mathrm{w}\mathrm{t}(a^j)-i_j-1)\ge -{\sum }_j{i}_j\ge \mathrm{\ell }(\mathrm{\ell }-1)\mathrm{/}2-\mathrm{\ell }(N-1)$. This quadratic has a finite lower bound and tends to positive infinity with $\mathrm{\ell }$. At fixed $d$ only finitely many lengths are possible. For each such length there are finitely many choices of the states in $A$; their total weight fixes the index sum, and the argument of step 3.1 gives finitely many index tuples. Therefore the full spanning family for $H_d(w)$ is finite; below the minimum of that quadratic it is empty. Finally the cyclic generation definition gives $W={\sum }_d{H}_d(w)$, without requiring directness.