General Hecke residue branching endpoints from normalized Schur cells

First construct the coefficient polynomials without specializing denominators. For $i<j$, $I=(i,j)$, $D\subseteq I$ and $p_D(t)=\max (\{i\}\cup \{d\in D:d<t\})$, define $H(A,B)={\sum }_{D\subseteq B\setminus A}(-1{)}^{\mathrm{\mid }D\mathrm{\mid }}\frac{{\prod }_{t\in A}(x_t-x_{p_D(t)})}{{\prod }_{t\in B}(x_t-x_{p_D(t)})}\mathrm{.}$ Cancel common $A\cap B$. For disjoint sets let $h=\min (A\cup B)$. If $h\in A$, multiply $H(A\setminus \{h\},B)$ by $x_h-x_i$. If $h\in B$, put $P=H(A,B\setminus \{h\})$; splitting the sum according to $h\in D$ gives $(P-P{\mathrm{\mid }}_{x_i=x_h})\mathrm{/}(x_h-x_i)$. Divided difference of $x_i^m$ is $-{\sum }_{a=0}^{m-1}{x}_i^a{x}_h^{m-1-a}$, so induction proves $H\in \mathbb{Z}[x]$ even at coincident coordinates. Also $H(\mathrm{\varnothing },B)={\delta }_{B,\mathrm{\varnothing }}$ by the same recursion. For $k\in A$, $h=\max ([i,k)\setminus A)$, splitting $x_k-x_{p_D(k)}=(x_k-x_h)+(x_h-x_{p_D(k)})$ gives $H(A,B)=(x_k-x_h)H(A\setminus \{k\},B)+{\mathbf{1}}_{h\mathrm{\ne }i}H((A\setminus \{k\})\cup \{h\},B)+{\mathbf{1}}_{k\in B}{H}_{i,k}(A\cap (i,k),B\cap (i,k))H_{k,j}(A\cap (k,j),B\cap (k,j))\mathrm{.}$ Indeed $(h,k)\subseteq A$, so common factors there first cancel. The second difference is zero when $h\in D$; otherwise its predecessor equals $p_D(h)$, giving the second term. The extra $k\in D$ terms have negative sign and split independently at $k$, giving the positive third product after subtraction. This accounts for every summand.

Put $C=I\setminus A$ and $K(A)={\sum }_{B\subseteq I}H(A,B){\prod }_{t\in \{i\}\cup B}(y_{t+1}-x_t)\mathrm{.}$ Expansion of each $y_{t+1}-x_{p_D(t+1)}$ as $(y_{t+1}-x_t)+(x_t-x_{p_D(t+1)})$ gives $K(A)={\sum }_{D\subseteq C}(-1{)}^{\mathrm{\mid }D\mathrm{\mid }}\frac{{\prod }_{t=i}^{j-1}(y_{t+1}-x_{p_D(t+1)})}{{\prod }_{t\in C}(x_t-x_{p_D(t)})}\mathrm{.}$ Selected difference factors vanish when their index belongs to $D$; the remaining predecessor is $p_D(t)$, and collection gives exactly $H(A,B)$ by cancellation of the complement numerator/denominator sets. For a bijection $\theta :C\to B_0\subseteq I$ with $\theta (c)\ge c$, this implies the matched-product identity $K(A)\equiv {\prod }_{t\in [i,j)\setminus B_0}(y_{t+1}-x_i)(\mathrm{m}\mathrm{o}\mathrm{d}(y_{\theta (c)+1}-x_c:c\in C))\mathrm{.}$ Here is an integral induction proof. If $C=\mathrm{\varnothing }$, the alternating expression is the asserted full product. Otherwise take $h=\min C$, remove its source and target, and apply induction on $(h,j)$. Splitting the alternating sum by $h\in D$ gives the prefix product times the divided difference of $K_{h,j}$ in $x_h$. The prefix is ${\prod }_{t=i}^{h-1}(y_{t+1}-x_i)$ and the divided difference is $(K_{h,j}{\mathrm{\mid }}_{x_h=x_i}-K_{h,j})\mathrm{/}(x_h-x_i)$. All remaining match generators are independent of $x_h$, so divided differencing retains their ideal. The induction product contains the target $\theta (h)$; modulo its match it vanishes before substitution, and after substitution its factor is $x_h-x_i$. Cancel that integral factor to leave exactly the tail product with this target removed. No target occurs below $h$, since targets are at least their sources. The prefix completes the stated product. All divisions here are polynomial divided differences, not evaluation of rational functions.

Suppose $q\mathrm{\ne }1$ and temporarily extend $F$ to a field $K$ containing $v$ with $v^2=q$. Take $N>d$, $W=W_N(\lambda )$, $L=W\mathrm{/}\mathrm{rad}{B}_{\tau }$, and $z=z_{\lambda }$. Use the actual normalized root operators ${\widehat{F}}_{i,j}=v^{-j}{K}_j{F}_{i,j}{K}_i{v}^{-i},c_{i,j}=\frac{v^{-2i-1}{K}_i^2-v^{-2j-1}{K}_j^2}{v-v^{-1}},b_{i,j}=\frac{v^{-2i-1}{K}_i^2-v^{-2j-1}{K}_{j+1}^2}{v-v^{-1}}\mathrm{.}$ The denominator is a field unit since $q\mathrm{\ne }1$. Define $S_{i,j}(A)={\sum }_{B\subseteq (i,j)}{\widehat{F}}_{i,b_1}\cdots {\widehat{F}}_{b_s,j}H(A,B)$, evaluating $x_k=-v^{-2k-1}{K}_k^2\mathrm{/}(v-v^{-1})$; use $S_{i,i}=1$. The recurrence of step 1.1 gives $S_{i,j}(A)=S_{i,j}(A\setminus \{k\})c_{h,k}+{\mathbf{1}}_{h\mathrm{\ne }i}{S}_{i,j}((A\setminus \{k\})\cup \{h\})+S_{i,k}(A\cap (i,k))S_{k,j}(A\cap (k,j))\mathrm{.}$ Left-interval toral coefficients commute with right-interval lower roots, so the product order is exactly as displayed. The empty-set operator is ${\widehat{F}}_{i,j}$. Rank-one commutation and the root recurrence give $E_i{\widehat{F}}_{i,i+1}={\widehat{F}}_{i,i+1}{E}_i+v{b}_{i,i}$, $E_i{\widehat{F}}_{i,j}=v^{-1}{\widehat{F}}_{i,j}{E}_i-v{\widehat{F}}_{i+1,j}$ if $j>i+1$, and $E_{j-1}{\widehat{F}}_{i,j}=v{\widehat{F}}_{i,j}{E}_{j-1}+{\widehat{F}}_{i,j-1}$ if $i<j-1$. Disjoint indices commute; adjacent exterior indices contribute precisely their $K$-weight factors. These follow by induction inserting $F_{i,j}=v{F}_{i+1,j}{F}_i-F_i{F}_{i+1,j}$ into the simple defining commutators.

Modulo the left ideal $U{E}_l$, the recurrence gives this complete table: $E_l{S}_{i,j}(A)=0$ if $l+1\in A$, or $l\mathrm{\notin }\{i\}\cup A$ and $l+1\mathrm{\notin }A\cup \{j\}$; it is $-v{S}_{i,l}(A\cap (i,l))S_{l+1,j}(A\cap (l+1,j))$ if $l\in \{i\}\cup A$, $l+1\mathrm{\notin }A\cup \{j\}$; it is $S_{i,j-1}(A)$ if $l=j-1\mathrm{\notin }\{i\}\cup A$; finally for $j-1\in A$, $h=\max ([i,j-1)\setminus A)$ it is $S_{i,j-1}(A\setminus \{j-1\})b_{h,j-1}+{\mathbf{1}}_{h\mathrm{\ne }i}{S}_{i,j-1}((A\setminus \{j-1\})\cup \{h\})$. To verify the table induct on ${\sum }_{a\in A}a$ using step 2.2. Empty $A$ is the elementary table. For $A=\{l+1\}$, $l>i$, expand again at $l$; the two cut remainders are ${\widehat{F}}_{i,l}{\widehat{F}}_{l+1,j}$ and its negative and cancel. At $l=i$ the cancellation is $b_{i,i}{\widehat{F}}_{i+1,j}={\widehat{F}}_{i+1,j}{c}_{i,i+1}$. For any other $k\in A$ the recurrence splits into intervals, the remote factor commutes, and the other factor uses induction. For the split row reduce to $l=i$ by expanding at $k=l$; if the replacement is $i+1$ its remainder is the vanishing row, and otherwise all terms combine into the recurrence for $S_{i+1,j}$. The shortened-endpoint row uses the same induction at the other end. Expanding at $k=j-1$ gives the final row: $E_{j-1}{\widehat{F}}_{j-1,j}$ contributes $v{b}_{j-1,j-1}$, its weight factor is $v^{-1}$, and $c_{h,j-1}+b_{j-1,j-1}=b_{h,j-1}$. These exhaust all positions. Successive commutation through a cut product gives $E_i\cdots E_{j-1}{\widehat{F}}^{B}z=v{\prod }_{t\in \{i\}\cup B}{b}_{t,t}(\lambda )z$. Summing proves $E_i\cdots E_{j-1}{S}_{i,j}(A)z=vK(A)(\lambda )z$.

Evaluate $b_{i,k}(\lambda )=(q^{{\lambda }_i-i}-q^{{\lambda }_{k+1}-k})\mathrm{/}(q-1)$ and $c_{i,k}(\lambda )=(q^{{\lambda }_i-i}-q^{{\lambda }_k-k})\mathrm{/}(q-1)$. Let $B_{i,j}$ be the zero-$b$ set in $[i,j)$ and $C_{i,j}$ the zero-$c$ set in $(i,j)$. Their vanishing is exactly the corresponding equal-residue comparison. If $c_{i,k+1}=0$, then $b_{i,k}\mathrm{\ne }0$, because $c_{i,k+1}-b_{i,k}=q^{{\lambda }_{k+1}-k-1}$ is a unit. Write $B\downarrow C$ for an injection of demands $B$ into supports $C$, sending each demand to a weakly smaller support. For $C\subseteq C_{i,j}$, we claim nonzero $S_{i,j}((i,j)\setminus C)z$ in $L$ implies $B_{i,j}\downarrow C$. Induct on $j-i$. At length one the elementary commutator and positive detection give equivalence with $b_{i,i}\mathrm{\ne }0$. At larger length some $E_l$ has nonzero image: otherwise all positive divided powers kill this vector, since its deficit has each simple-root coefficient zero or one, so powers greater than one have absent target weights; triangular spanning would generate a proper second highest submodule of the simple $L$. Use the table of step 3.1. In the split case its disjoint factors commute, so both their vectors are nonzero; induction gives matchings on both intervals, and $c_{i,l+1}=0$, $b_{i,l}\mathrm{\ne }0$ identify the split demands with the full ones. If $j-1\in C$, induction on the shortened interval matches old demands and the extra demand, when present, uses $j-1$. Otherwise put $h=\max (\{i\}\cup C)$ in the last table row. A nonzero first term has $b_{h,j-1}=b_{i,j-1}\mathrm{\ne }0$, so there is no new demand. A nonzero second term matches old demands into $C\setminus \{h\}$; the released largest support $h$ handles the only possible new demand $j-1$. This proves the claim in every case.

A primitive line of $L^1$ at $\lambda (i)$ exists exactly when $B_{i,N}\downarrow C_{i,N}$. For necessity, dualize a nonzero primitive map using [F1] to get $L^1\to \mathrm{\nabla }(\lambda (i))$. Its highest coefficient functional on $W^1$ is nonzero at ${\widehat{F}}_{i,N}z$ and zero at all nonempty cut products, by [F1]'s exact coefficient test. On $S_{i,N}(A)z$, $A=(i,N)\setminus C_{i,N}$, its value is therefore that nonzero value times $H(A,\mathrm{\varnothing })(\lambda )={\prod }_{t\in A}{c}_{i,t}(\lambda )\mathrm{\ne }0$. This vector survives the radical, so step 4.1 gives necessity. Conversely choose a matching with image $C\subseteq C_{i,N}$ of size $\mathrm{\mid }{B}_{i,N}\mathrm{\mid }$ and invert it to a bijection $C\to B_{i,N}$ with targets at least their sources. The polynomial match generators in step 2.1 vanish, so $K((i,N)\setminus C)(\lambda )={\prod }_{t\in [i,N)\setminus B_{i,N}}{b}_{i,t}(\lambda )\mathrm{\ne }0$. Step 3.1 makes the resulting $w$ nonzero in $L$. Its lower-Levi commutators vanish in $L$: a potentially nonzero split has $l+1\in C$ and demands $B_{i,l}\cup B_{l+1,N}$. If its first factor survives, step 4.1 gives $\mathrm{\mid }{B}_{i,l}\mathrm{\mid }\le \mathrm{\mid }C\cap (i,l)\mathrm{\mid }$; since $\mathrm{\mid }{B}_{i,N}\mathrm{\mid }=\mathrm{\mid }C\mathrm{\mid }$ and $l+1\in C$, the suffix demands outnumber suffix supports, forcing the second factor into the radical. Higher divided generators have absent target weights. Thus $w$ is primitive. [F1] gives one-dimensionality and excludes every other primitive highest label.

Let $v_i$ be the nonzero primitive line for a matched removable row $i$ and ${\mathrm{\Psi }}_i$ its cyclic lower-Schur module. If it has a proper nonzero submodule, that submodule contains a simple submodule, whose highest line is $v_h$ by [F1] and step 5.1. Triangular spanning puts its weight strictly below $\lambda (i)$, so $h<i$; equality would contain the generating line. The universal Weyl map onto ${\mathrm{\Psi }}_i$ and [F2] force $\lambda (h)$ and $\lambda (i)$ to have the same residue multiset, hence $c_{h,i}=0$. Conversely suppose $h<i$ are matched and $c_{h,i}=0$. The demand/support sets split as $B_{h,N}=B_{h,i}\cup B_{i,N}$ and $C_{h,N}=C_{h,i}\cup \{i\}\cup C_{i,N}$. A prefix matching exists by restricting any matching for $h$ to demands below $i$; these cannot use later supports. Combine it with a matching for $i$, giving an image $C$ omitting $i$ with exactly $\mathrm{\mid }{B}_{h,i}\mathrm{\mid }$ earlier and $\mathrm{\mid }{B}_{i,N}\mathrm{\mid }$ later supports. Let $A=(h,N)\setminus C$, $k=\max (\{h\}\cup (C\cap (h,i)))$. Then $c_{k,i}=0$, and the recurrence at $i$ gives $S_{h,N}(A)z={\mathbf{1}}_{k\mathrm{\ne }h}{S}_{h,N}((A\setminus \{i\})\cup \{k\})z+S_{h,i}(A\cap (h,i))S_{i,N}(A\cap (i,N))z\mathrm{.}$ When present, the first term replaces support $k<i$ by $i$; its earlier demands now outnumber earlier supports, so step 4.1 kills it in $L$. The other full and suffix vectors are the nonzero primitive vectors of step 5.1. Hence $v_h$ lies in ${\mathrm{\Psi }}_i$ and generates a proper submodule. Thus ${\mathrm{\Psi }}_i$ is simple precisely for the first matched row of its residue, and the full first-level socle is the multiplicity-free sum of these simples.

The matching criterion is exactly the signature criterion. A zero $c_{i,k}$ records a later removable push of the same residue and a zero $b_{i,k}$ a later addable pop; nonremovable equal rows have both slots and cancel as a neutral pair. Removing common demand/support slots preserves the prefix inequalities for $B\downarrow C$. The remaining prefix inequalities say that the minus at row $i$ survives all later pops. Thus matched rows are normal, and the first matched row of a residue is its good row. Good deletion preserves restrictedness: its only possible new gap $e$ is above the deleted row, when ${\lambda }_{i-1}-{\lambda }_i=e-1$. The adjacent removable corners then have the same residue and no intervening same-residue boundary corner; if the lower minus survives, the upper minus survives too, contradicting first-surviving selection. The dual adjacent-plus argument proves cogood addition preserves restrictedness. Flipping the selected minus to plus changes no other $i$-boundary status, since all other changed corners have residues $i\pm 1$, different from $i$ even at $e=2$. Before the first surviving minus the stack is empty, and every suffix prefix has at least as many remaining minus pushes as plus pops; after the flip these are exactly the conditions for the last surviving plus. Hence good deletion and cogood addition are inverse partial moves on restricted partitions.

Transfer the socle through the actual corner, using more than exactness. By steps 6.1 and 7.1 all Schur socle labels of $V=L_N(\lambda {)}^1$ are restricted; [F3] makes their lower corners nonzero simples. Therefore $V$ has no nonzero corner-killed submodule, since every nonzero finite-dimensional submodule contains a simple submodule. For a simple lower-Hecke injection $D\hookrightarrow e_{\mathrm{l}\mathrm{o}\mathrm{w}\mathrm{e}\mathrm{r}}V$, the corner adjunction gives $A{e}_{\mathrm{l}\mathrm{o}\mathrm{w}\mathrm{e}\mathrm{r}}{\otimes }_{e_{\mathrm{l}\mathrm{o}\mathrm{w}\mathrm{e}\mathrm{r}}A{e}_{\mathrm{l}\mathrm{o}\mathrm{w}\mathrm{e}\mathrm{r}}}D\to V$, whose image $I$ is generated by $D$ and satisfies $e_{\mathrm{l}\mathrm{o}\mathrm{w}\mathrm{e}\mathrm{r}}I=D$. A proper nonzero submodule $J<I$ has corner zero or $D$. The first contradicts the no-killed-submodule property; the second contains $D$ and thus generates $I$. Hence $I$ is simple. Conversely every surviving Schur socle simple has simple corner by [F3]. These inclusions prove ${\mathrm{soc}}_H(e_{\mathrm{l}\mathrm{o}\mathrm{w}\mathrm{e}\mathrm{r}}V)=e_{\mathrm{l}\mathrm{o}\mathrm{w}\mathrm{e}\mathrm{r}}{\mathrm{soc}}_A(V)$, with multiplicities and actual labels. [F1]'s explicit corner identifies this with the actual Hecke restriction of $D(\lambda )$.

Define the primary functors over $F$. By [F3] the spectra of $L_d$ in the regular module are q-contents, hence among the distinct $[i{]}_q$, $i\in \mathbb{Z}\mathrm{/}e$; cellular layers and their common triangular flags give this for the regular action as well. The polynomial CRT applied to ${\prod }_i(X-[i{]}_q{)}^{d!}$ gives mutually orthogonal projectors $P_{d,i}$, with sum one, in $F[L_d]$. They centralize $H_{d-1}$ and are star-fixed. Taking exponent $d!$ is sufficient for every Jordan block of the regular action and therefore every module; absent eigenvalues give the zero projector. On a good lower simple $D(\lambda \setminus A)$ in restriction, $L_d={\sum }_{r\le d}{L}_r-{\sum }_{r<d}{L}_r$ has scalar $[c(A){]}_q$: the symmetric sums have their actual cell scalars by [F2]. Thus its primary projector selects exactly the good residue. Every nonzero primary summand contains a simple socle submodule, so a summand with no good residue is zero. The form quotient makes $D(\lambda )$ star self-dual, and its star-fixed projector inherits a nondegenerate restricted form; its head is dual to its simple socle and has the same label. This proves the restriction endpoints when $q\mathrm{\ne }1$. At $q=1$ use the actual group endpoint [F4], with the same primary construction and residue normalization.

Induction and restriction are biadjoint over $F$. Let $B=H_{d+1}$, $H=H_d$, and let $E_H:B\to H$ retain precisely the lower standard-basis terms. It is an $H$-bimodule map. For distinguished parabolic cosets $c,t$, the perfect trace gives $E_H(T_c^{\ast }{T}_t)={\delta }_{c,t}{q}^{\mathrm{\ell }(t)}$: pair against each lower basis element using ${\tau }_B(h{E}_H(a))={\tau }_B(ha)$ and the disjoint length-additive coset supports. All diagonal values are units. Hence the relative Frobenius pairing gives $B{\otimes }_{H}X\cong {\mathrm{Hom}}_H(B,X)$ naturally. For a commuting primary projector $p=P_{d+1,i}$, cyclicity of the trace and commutation with $H$ give $E_H(pa)=E_H(ap)$, by pairing both expressions against the lower algebra's perfect trace. Under this identification right multiplication by $p$ corresponds to selecting $p$ on restriction. Ordinary tensor adjunction and this Frobenius adjunction therefore give both ${\mathrm{Hom}}_B(F_{i}X,Y)={\mathrm{Hom}}_H(X,E_{i}Y)$ and ${\mathrm{Hom}}_B(Y,F_{i}X)={\mathrm{Hom}}_H(E_{i}Y,X)$.

For every simple label $\eta$ of rank $d+1$, step 9.1 and absolute simplicity give each of these Hom dimensions with $X=D(\mu )$, $Y=D(\eta )$ as one exactly when the good $i$-deletion of $\eta$ is $\mu$, and zero otherwise. Step 7.1 makes its inverse exactly the cogood addition, unique when defined. The complete cellular classifier [F3] covers every simple, so these two Hom tests identify respectively the simple head and simple socle of $F_{i}D(\mu )$, once each. If no cogood addition exists both tests vanish for every simple, forcing $F_{i}D(\mu )=0$ by finite dimensionality. At rank zero $H_0=H_1=F$, $L_1=0$, and the only move adds the residue-zero node; these formulas give $F_{0}F=F$ and every other summand zero.

Descend the temporary square-root extension. The Hecke modules, their cell forms, all primary projectors and functors above are defined over $F$. The cellular quotients of [F3] remain absolutely simple under $F\to K$, and finite-dimensional Hom spaces are kernels of finite linear systems over $F$, so their dimensions are unchanged by this field extension. Step 9.1's restriction and step 11.1's induction Hom tests therefore hold over $F$ itself and give exactly the stated endpoints. This argument also permits a purely inseparable square-root extension in characteristic two; no Galois averaging is used. The polynomial, integral module, actual corner and coefficient-functional constructions never divide a specialized Gram determinant. No primality condition on $e$ or Fock-space label classification entered the proof; no Choice principle is used.