Signed forest candidates, wall homomorphisms, and their composition

The hook quotient is a polynomial by induction on its forest. Removing the root of a tree cancels its factor $[\mathrm{\mid }F\mathrm{\mid }{]}_T$, leaving the quotient for its subforest. For a disjoint union of forests of sizes $u,v$, the quotient is $B_{u+v,u}(T)$ times their two quotients, a polynomial by [F2]. The empty forest has weight $1$. For a chain, numerator and denominator cancel completely, so its weight is $1$. All further calculations may therefore be proved in $\mathbb{Q}(T)$ and then interpreted as identities in $\mathbb{Z}[T]$, before making any field specialization.

We first use loop value $-T$. Consider matchings with a fixed finite number $d$ of cups on a half-infinite ordered row, with all sufficiently far-left sites propagating. A diagram is encoded by the $d$ right cup endpoints, numbered from the right starting at $1$: scanning from right to left, a listed site opens a cup and the next unpaired unlisted site closes the most recent cup; unlisted sites with no open cup propagate. Thus every $d$-element subset of the positive integers encodes exactly one matching. Hook weights are unchanged by adding more leftmost propagating strands, because bending one more such strand adds a root above the entire old forest and cancels its new numerator factor. We may consequently form the formal series $V_d$ of these matchings with their hook weights. Every fixed output coefficient after a local generator or cup attachment involves finitely many terms: only its finitely many cups and the nearest exposed propagating strand can be reconnected. All operations below are defined coefficient by coefficient.

We prove that every local generator kills $V_d$. Fix an output matching $D$ with its simple cup at sites $i,i+1$, and remove that cup to obtain a gap $J$ in $D^{-}$. A line is exposed to $J$ if an arc from $J$ to an interior point of that line can avoid all other lines. Replacing that line by two lines ending at $i$ and $i+1$ is its nip at $J$. Exactly $D$ itself and these nipped diagrams are the preimages contributing to the coefficient of $D$: a generator changes only the two lines meeting its sites, while the converse reconnection recovers each nip. A line exposed on both sides is nipped only once. Therefore the coefficient is $C_D=-[2{]}_{T}h(D)+{\sum }_{e\text{ exposed to }J}h({\mathrm{nip}}_J(e))\mathrm{.}$ This enumeration includes loops through the first term and all reconnection terms with coefficient $1$.

Here is a textual reduction verifying this coefficient identity. First contract each intervening fixed internal subforest to a weighted interval block: its total edge count is retained in the containing arc's subtree size and its own hook factors are common to all compared terms. Consecutive exposed arcs then have adjacent block endpoints; relabel these endpoints in their order without changing subtree sizes. Suppose those two arcs have endpoints $(a,b),(b+1,c)$, with all their internal subforests retained, and the distinguished simple cup is immediately beyond them at $(c+1,c+2)$. Replace those two arcs by $(a,c),(b,b+1)$. If their subtree sizes are $E_1,E_2$, their merged subtree has size $E_1+E_2$. Every term in $C_D$ other than the two nips of these arcs corresponds to the replaced diagram with hook ratio $[E_1+E_2{]}_T\mathrm{/}([E_1{]}_T[E_2{]}_T)$. Comparing the two exceptional nips with the single exceptional nip after replacement leaves a common rational factor times $[E_2{]}_T[E_1+E_2+1{]}_T+[E_1{]}_T-[E_2+1{]}_T[E_1+E_2{]}_T=0.$ To verify the ratios, every unchanged arc contributes the same denominator factor; the changed arcs contribute respectively the subtree sizes $E_1,E_2,E_1+E_2,E_2+1,E_1+E_2+1$. The displayed identity follows by induction on $E_1$ from the recurrence for $[j{]}_T$, with $E_1=0,1$ as immediate base cases. Thus this replacement preserves the vanishing of $C_D$. The reflected replacement does the same on the other side. Each replacement reduces by one the number of exposed arcs on the chosen side; repeated replacements leave at most one exposed arc on each side.

In the half-infinite setting there is always a leftmost propagating strand beyond all the cups under consideration. Bend that strand and the other output strands to put the distinguished cup and the remaining exposed arcs inside one outer arc. This only adds successive roots as in step 2.1. After the reductions of step 4.1, let the sizes of the optional left and right exposed subtrees be $B,C>0$. Canceling the hook factors of unchanged internal subforests, the four contributions in step 3.1 have the common-factor-normalized sum $-\frac{[2{]}_T}{[B{]}_T[C{]}_T[B+C+2{]}_T}+\frac{1}{[B+1{]}_T[C+1{]}_T[B{]}_T[C{]}_T}+\frac{1}{[B+1{]}_T[C{]}_T[B+C+2{]}_T}+\frac{1}{[B{]}_T[C+1{]}_T[B+C+2{]}_T}=0.$ The numerator is $-[2{]}_T[B+1{]}_T[C+1{]}_T+[B+C+2{]}_T+[B{]}_T[C+1{]}_T+[C{]}_T[B+1{]}_T$, zero by the same recurrence. If the left subtree is absent, the three terms reduce instead to $-[2{]}_T\mathrm{/}([C{]}_T[C+2{]}_T)+1\mathrm{/}([C+1{]}_T[C{]}_T)+1\mathrm{/}([C+1{]}_T[C+2{]}_T)$, zero because $[C+2{]}_T+[C{]}_T=[2{]}_T[C+1{]}_T$. The case of no right subtree is its reflection. If neither is present, the sum is $-[2{]}_T\mathrm{/}[2{]}_T+1=0$. Thus $C_D=0$ in every case, proving the assertion of step 3.1 as a polynomial identity.

A formal vector killed by every generator is uniquely determined by its coefficient on the fully nested rightmost $d$ cups. Indeed order the endpoint subsets componentwise, with minimum $\{1,\dots ,d\}$. For any other subset $J$, let $i>1$ be its least element with $i-1\mathrm{\notin }J$, and replace $i$ by $i-1$ to obtain $K<J$. In the coefficient at $K$ after applying the generator at $i-1,i$, the coefficient at $J$ occurs with coefficient $1$. Every other contributing subset $L$ satisfies $L<J$: nipping a line different from the one producing $J$ moves an endpoint closer to the right boundary, and the loop term is $K<J$. Setting this coefficient to zero therefore determines the coefficient at $J$ from smaller subsets. Componentwise lower sets are finite, so induction applies. This proof is valid over $\mathbb{Z}[T]$ and after any specialization.

We check when the finite candidate $v_{a,b}$ is killed by every generator. Put $d=(a-b)\mathrm{/}2$ and $M=(a+b)\mathrm{/}2+1$. The cancellation of step 3.1, step 4.1, step 5.1 applies unchanged whenever an enclosing arc or a propagating strand is exposed to the left of the distinguished cup. Only the remaining case has a missing nip: the distinguished cup is not enclosed by another cup, and there is no propagating strand to its left. Number the original input sites $1,\dots ,a$ and let the distinguished cup be $(i,i+1)$. Then every site $1,\dots ,i-1$ is paired within that interval, so $i$ is odd. Add a new leftmost input site $0$ and a new first output site, joined by a propagating strand. Remove $(i,i+1)$ and nip that new strand at its gap, producing the cup $(0,i)$ and a strand from $i+1$ to the new output site. This is the unique additional contributing diagram. Its bent matching has $(a+1+b+1)\mathrm{/}2=M$ edges. The new cup $(0,i)$ and the $(i-1)\mathrm{/}2$ original cups inside it form a separate tree with $t=(i+1)\mathrm{/}2$ edges. Since the original diagram had these $t-1$ cups and the distinguished cup, $1\le t\le d$. Every other edge lies to its right, including the bent strand beginning at $i+1$, so no edge of this tree is comparable with an edge of the remaining forest. Any common roots introduced by auxiliary bending of additional strands cancel as in step 2.1 before this split is counted. By step 1.1 its hook weight consequently factors as $B_{M,t}(T)$ times the polynomial hook weights of those two parts. The uncanceled coefficient is the negative of this missing hook weight. For $a=3,b=1,i=1$, the missing matching has a one-edge component and a two-edge chain, giving $B_{3,1}(T)=[3{]}_T$, whose negative agrees with the direct coefficient $1-T^2$.

For a half-infinite basis diagram $E$ with $r$ cups, define its attachment with $s-r$ additional cups to be $E$ composed with the formal hook vector ${\mathsf{V}}_{s-r}$ of step 2.1; coefficientwise finiteness was established there. Write ${\mathrm{Att}}_{r,s}(E)$ for this attachment. For a generator whose two sites stay on one side of the attachment seam, associativity matches each coefficient in the generator acting on ${\mathrm{Att}}_{r,s}(E)$ with the corresponding coefficient in ${\mathrm{Att}}_{r,s}$ applied after the generator acts on $E$; any loop has factor $-T$ in both diagrams. In the remaining case the generator joins two propagating strands across the seam. Fix an output matching $D$ and delete the common unchanged arcs away from its newly formed simple cup gap $J$. Their hook factors are a common factor $H_{\mathrm{o}\mathrm{u}\mathrm{t}}$. In the remaining local matching $D_0$, the preimages contributing to its coefficient are exactly $D_0$ and the diagrams obtained by nipping one line exposed to $J$; a line exposed from both sides is counted once. Thus the difference of the two coefficients is $H_{\mathrm{o}\mathrm{u}\mathrm{t}}$ times the seam coefficient $C_{D_0}=-[2{]}_T{h}_{F(D_0)}(T)+{\sum }_{e\text{ exposed to }J}{h}_{F({\mathrm{nip}}_J(e))}(T)\mathrm{.}$ The first term is the loop contribution, since the loop value is $-T=-[2{]}_T$, and every nip contributes its reconnection with coefficient $1$. All diagrams in this sum have the same number of arcs. Every arc outside the smallest region changed by a nip has the same subtree size in each term, so its hook denominator is a common factor; the factorial numerator is common as well. After cancelling these factors, the remaining coefficient is exactly the exposed-arc calculation of step 4.1, step 5.1: adjacent exposed arcs reduce to at most one on either side, and the resulting four-term expression (or its one-sided or empty version) vanishes by the displayed quantum-integer recurrence identities there. Hence $C_{D_0}=0$. This proves that attachment commutes with every generator, including one joining two propagating strands. By coefficientwise linearity, $U_i{\mathrm{Att}}_{r,s}(v)={\mathrm{Att}}_{r,s}(U_{i}v)$ for each formal input $v$. In particular, attachment carries an input killed by every generator to an output killed by every generator.

Composing attachments with $u=(a-b)\mathrm{/}2$ and $v=(b-c)\mathrm{/}2$ cups gives a formal vector with $u+v$ cups. The first attachment applied to the zero-cup identity produces ${\mathsf{V}}_u$, which is killed by every generator by step 5.1. Applying step 6.3 to this annihilated input makes the composite killed by every generator; by step 6.1 it is a scalar multiple of the formal vector with $u+v$ cups. To find the scalar, take the fully nested monic diagram $M:a\to c$, whose bent forest is a chain and whose coefficient is $1$. It has a unique monic factorization $M=Y\circ X$ through $b$: each cap of $X$ is a cap of $M$, and no cap of $X$ can contain a through strand that $Y$ later joins, so the $u$ caps of $X$ are exactly the $u$ innermost caps of $M$; the remaining $v$ caps belong to $Y$. After bending, $F(X)$ consists of a common chain of $v+c$ roots above two disjoint chains of lengths $u$ and $v$, while $F(Y)$ and $F(M)$ are chains. Therefore $h_{F(X)}(T)=\frac{[u+2v+c{]}_T!}{({\prod }_{j=u+v+1}^{u+2v+c}[j{]}_T)[u{]}_T![v{]}_T!}=\frac{[u+v{]}_T!}{[u{]}_T![v{]}_T!}=B_{u+v,u}(T),$ and $h_{F(Y)}(T)=h_{F(M)}(T)=1$. Monic composition creates no closed component, so the coefficient of $M$ in the composite is exactly $B_{u+v,u}(T)$. This is the scalar multiplying the formal vector. Restricting to a finite row and discarding diagrams that use a site beyond its left boundary preserves composition, since further cup insertion cannot remove such a site. With the stated $u,v$, the finite candidate identity follows.

Now specialize $T=-\delta$, so the loop value $-T$ is the required $\delta$. The recurrence gives $[j{]}_{-\delta }=(-1{)}^{j-1}[j{]}_{\delta }$ by induction. Thus the quantum characteristic and all mixed-radix zero tests are unchanged. The hypotheses give $0<d<s_k$ and $s_k\mid M$. For every $1\le t\le d$, at least one digit of $t$ below place $k$ is positive, while every digit of $M$ at those places is zero. By [F2], $B_{M,t}(-\delta )=0$. Step 6.2 therefore kills every finite failure coefficient. All generators annihilate the candidate. Its chain coefficient is $1$ by step 1.1, so it is nonzero.

Finally, if a linear combination $v\in W_{a,b}$ is killed by all generators, every adjacent cap at the $a$ inputs kills it modulo propagation less than $b$. This follows by factoring the generator as cup followed by cap: insertion of that cup is injective on the relevant monic diagram basis, so zero after insertion implies zero before it. Every diagram $n\to a$ of propagation less than $a$ has a cup on its $a$-boundary, and an innermost such cup is adjacent; factoring it off shows that its postcomposition with $v$ is zero modulo propagation less than $b$. Postcomposition consequently descends from the quotient $W_{n,a}$ to $W_{n,b}$ and commutes with the diagram action. Nonzeroness for every $n\ge a$ follows by choosing a monic diagram $z:n\to a$ with a left retraction, by [F4]: because $a>b\ge 0$, one has $a>0$, and the shifted adjacent-cup construction gives this retraction over every field without division. Composing a putative zero $zv$ with the retraction would give $v=0$, a contradiction. Step 7.2 supplies precisely this annihilated nonzero $v$ under the stated wall hypotheses. Together with step 6.3, step 7.1 and the substitution in step 7.2, this proves all conclusions.