An integral essential-subset basis and count for type-A root intervals

Extend only to $K=F(t)$. This field is infinite, and extension preserves the rank of every finite Gram matrix. A nonzero Schur cell-form quotient stays simple after this extension: in any field a vector outside its radical generates the entire cell module by the cellular rank-one action, and the same argument applies to the extended nonzero Gram form. No algebraic closure or infinite-choice step is required.

At $q=1$, identify ${⨁}_{\nu }{M}^{\nu }$ with the tensorword basis of $(K^N{)}^{\otimes m},m=\mathrm{\mid }\mu \mathrm{\mid }$, by sending $m_{\nu }{T}_d$ to the word whose $j$th letter is the row containing entry $j$ in the row-standard tableau $t^{\nu }d$. This is the explicit right-coset/tensorword convention; for regular type and basis $d$, that letter is $j{d}^{-1}$ under right permutation notation. The symmetric-group action on tensor positions is transported through this map, so inverses in labels and right-position actions are fixed consistently. The symmetric group permutes the tensor positions. An endomorphism commuting with this permutation action has matrix coefficients constant on its orbits on pairs of words. Such orbits are indexed by nonnegative integer matrices $A=(A_{ab})$ with $\sum A_{ab}=m$, recording the number of positions with input $b$ and output $a$. Let $E_A$ be the matrix with entry 1 on that orbit and 0 elsewhere. The $E_A$ have disjoint support and form a basis of the endomorphism algebra. For a matrix $g$, $g^{\otimes m}={\sum }_A({\prod }_{ab}{g}_{ab}^{A_{ab}})E_A$. The transpose involution sends $E_A$ to $E_{A^T}$; it is precisely the $q=1$ Schur star (the corresponding semistandard Hom maps are transposes under the orthonormal tensorword/permutation forms).

Invertible-matrix evaluations span this Schur algebra over $K$. Indeed a linear functional annihilating all $g^{\otimes m}$ gives a polynomial $P(g)={\sum }_A{c}_A{g}^A$ vanishing whenever $\det (g)\mathrm{\ne }0$. Then $P\det$ vanishes at every matrix over $K$. An elementary induction on the number of variables shows that a polynomial over an infinite field vanishing everywhere is zero: regard it as a polynomial in the last variable and use infinitely many values, then apply the induction to its coefficients. Thus $P\det =0$ as a polynomial, and the polynomial domain gives $P=0$, so all $c_A$ vanish. Finite-dimensional duality proves the spanning claim. Consequently the star-converted Schur simple is a simple rational left ${\mathrm{GL}}_N$-module, with the same submodules and composition multiplicities as the right Schur module.

The canonical semistandard tableau has weight $\mu$ and Schur form value 1. Indeed its diagonal Schur basis element is the projection onto the permutation summand $M^{\mu }$: the defining Hom map is the identity there and zero on the other summands. Its square is itself, so the defining cell-product coefficient, which is the canonical tableau norm, is 1. This argument uses the canonical diagonal projection only, not an idempotence assertion for all diagonal tableau elements. Every semistandard tableau weight $\nu$ is dominated by $\mu$: for each $k$, all entries $\le k$ occur in the first $k$ rows by column strictness, so there are at most ${\mu }_1+\cdots +{\mu }_k$ of them. The unique tableau of weight $\mu$ is the canonical one. Thus $L$ has highest weight $\mu$ with one-dimensional highest space $Kv$, and its positive upper unipotent group fixes $v$: any nonconstant positive-root coefficient would have weight strictly above $\mu$. Torus weights are integer characters, distinct as Laurent monomials over $K$; no Lie-eigenvalue congruence substitutes for weight equality.

On the open set where leading principal minors are nonzero, Gaussian elimination gives $g=u^{-}d{u}^{+}$. The negative-unipotent coefficient operators on $v$ span $L$. To see this, let $W$ be their span and let $\mathrm{\ell }$ annihilate $W$. Then $\mathrm{\ell }(gv)=0$ on that open set because $u^{+}v=v$ and $dv$ is a scalar multiple of $v$. A polynomial matrix coefficient vanishing there vanishes identically by the same polynomial-identity argument, so $\mathrm{\ell }$ annihilates the coefficient orbit of $v$ under all ${\mathrm{GL}}_N$. That orbit spans $L$ by simplicity. Hence $W=L$.

Positive coefficient operators detect every nonzero vector $x$ of weight $\mu -\gamma$. If they all had zero highest coefficient on $x$, the Gaussian factorization would imply that the highest coefficient of $gx$ was zero on the same open set: negative operators cannot raise a weight to $\mu$, and no weight exceeds $\mu$. The resulting polynomial identity would make that coefficient zero on the whole simple-module orbit of $x$, which spans $L$ and contains $v$, a contradiction. Thus some positive coefficient operator of weight $\gamma$ raises $x$ to a nonzero multiple of $v$.

For $\gamma =\alpha (r,s)$, every simple-root coefficient is 1. In a product expansion of a unipotent matrix, any monomial of total root weight $\gamma$ therefore has each root-variable exponent $0$ or $1$, and its roots are pairwise disjoint intervals covering $[r,s]$. They are exactly cut partitions of that interval. Put $f_{a,b}=E_{b+1,a},e_i=E_{i,i+1},Y_{r,s}(U)=f_{r,u_1-1}{f}_{u_1,u_2-1}\cdots f_{u_k,s}$, for $U=\{u_1<\cdots <u_k\}\subset (r,s]$, with $Y(\mathrm{\varnothing })=f_{r,s}$. These words span the required negative coefficient space. Similarly the positive coefficient space is spanned by words using each $e_r,\cdots ,e_s$ once. Root-matrix commutators expand every longer positive root into such simple words. Permutations of these simple letters are identified using $e_i{e}_j=e_j{e}_i$ for $\mathrm{\mid }i-j\mathrm{\mid }>1$; their relative orders on adjacent pairs determine the canonical representative $Q_{r,s}(I)=e_{i_k}\cdots e_{i_1}\cdot {\prod }_{j\in [r,s]\setminus I}^{\mathrm{i}\mathrm{n}\mathrm{c}\mathrm{r}\mathrm{e}\mathrm{a}\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{g}}{e}_j,I=\{i_1<\cdots <i_k\}\subset (r,s]$. Every word is carried to this representative by exchanging adjacent incomparable letters in the corresponding path partial order. Hence these $Q$ words span all positive tests at the interval weight. There are no divided powers greater than 1 or factorial denominators in this grade.

The actions and all commutator calculations below use $[E_{ab},E_{cd}]={\delta }_{bc}{E}_{ad}-{\delta }_{da}{E}_{cb},e_{i}v=0$, and $(E_{ii}-E_{i+1,i+1})v=a_{i}v$. They are integral matrix identities and apply to the actual Schur module. The construction descends from $K$ to $F$ because the resulting matrices have entries in $F$ and scalar extension preserves all ranks and linear relations.

For $M,U\subset (r,s]$, define ${\zeta }_{r,s}(M,U)$ recursively. For $M=\mathrm{\varnothing },\zeta (\mathrm{\varnothing },U)={\prod }_{j\in (r,s]\setminus U}C(r,j)$. For $M\mathrm{\ne }\mathrm{\varnothing }$ let $m=\min M$ and $M^{\mathrm{\prime }}=M\setminus \{m\}$. If $m\mathrm{\notin }U$, put $\zeta (M,U)=\zeta (M^{\mathrm{\prime }},U)\mathrm{/}C(r,m)$. If $m\in U$, put $\zeta (M,U)=[\zeta (M^{\mathrm{\prime }},U)-{\zeta }_{r,m-1}(\mathrm{\varnothing },U\cap (r,m)){\zeta }_{m,s}(M^{\mathrm{\prime }},U\cap (m,s])]\mathrm{/}C(r,m)$. Every division here is performed in the integer polynomial ring before specialization. It is exact. To prove this, induct on $\mathrm{\mid }M\mathrm{\mid }$, keeping the stronger statement that $\zeta (M,U)={\zeta }_{r,m-1}(\mathrm{\varnothing },U\cap (r,m))P(C(r,m))$, where $P$ is an integer polynomial in the remaining differences independent of $r$ and of $U\cap (r,m)$. At the next smallest cut, use $C(r,j)=C(r,m)+C(m,j)$. If $m\mathrm{\notin }U$, the numerator contains the explicit factor $C(r,m)$. If $m\in U$, the two terms are respectively evaluations $P(C+D)$ and $P(D)$, multiplied by the same left factor, with $C=C(r,m)$; their difference is divisible by $C$, since $(C+D{)}^h-D^h=C{\sum }_{a=1}^h\left(\genfrac{}{}{0px}{}{h}{a}\right)C^{a-1}{D}^{h-a}$. This gives the stronger factorization at the next step. The base case is the displayed product. Thus all coefficients are integer polynomials even when their displayed denominators vanish modulo $p$.

Define $T_{r,s}(M)={\sum }_U{\zeta }_{r,s}(M,U)Y_{r,s}(U)$. The coefficient recursion is equivalently the exact negative-word recursion $C(r,m)T_{r,s}(M)=T_{r,s}(M^{\mathrm{\prime }})-T_{r,m-1}(\mathrm{\varnothing })T_{m,s}(M^{\mathrm{\prime }})$. $(T)$ This follows by separating cut words according to whether the cut $m$ occurs. Products of adjacent interval words concatenate with the boundary cut present; this is exactly the cut-word product in the numerator.

The following splitting identity is integral: $T_{r,i-1}(E)T_{i,s}(F)=T_{r,s}(E\cup F)-T_{r,s}((E\setminus \{e\})\cup \{i\}\cup F)-C(e,i)T_{r,s}(E\cup \{i\}\cup F),(P)$ where $E\subset (r,i),F\subset (i,s],e=\max E$. If $E=\mathrm{\varnothing }$ omit the middle term and set $e=r$. For $E=\mathrm{\varnothing }$ this is $(T)$ rearranged at cut $i$. For nonempty $E$, remove its smallest $m$, insert $(T)$ on the left and apply the induction to the two shorter products. Use $C(r,i)=C(r,m)+C(m,i)$. The terms containing the extra cut $m$ cancel in pairs, and the remaining numerator is $C(r,m)[T(E\cup F)-T((E\setminus \{e\})\cup \{i\}\cup F)-C(e,i)T(E\cup \{i\}\cup F)]$. Exact divisibility already proved permits cancellation in the integer polynomial ring; reduction to $F$ is made only afterwards. This proves $(P)$ by induction, including the singleton $E$ case.

Order subsets by their largest differing element, with the subset containing that element larger. Complements reverse this order. The transition from $T(M)$ to $Y(U)$ is unit triangular: $T_{r,s}(M)=Y_{r,s}((r,s]\setminus M)+{\sum }_{U<(r,s]\setminus M}{c}_U{Y}_{r,s}(U)$. $(U)$ Here is an explicit induction. First $(T)$, descending on $j$, gives $T_{r,s}([j,s])={\sum }_{U\subset (r,j)}({\prod }_{d\in (r,j)\setminus U}C(r,d))Y_{r,s}(U)$. For $j=s+1$ this is the defining empty-$M$ product; to pass to $j$ subtract $T_{r,j-1}(\mathrm{\varnothing })T_{j,s}([j+1,s])$ in $(T)$. The terms with cut $j$ cancel, leaving the indicated sum divided by the common factor $C(r,j)$. At $j=r+1$ this gives $T((r,s])=Y(\mathrm{\varnothing })$. For any other $M$ choose $i\mathrm{\notin }M$, split $M=E\cup F$ around $i$, and rearrange $(P)$. The product of the two shorter leading $Y$ words is the cut word for the complement of $M$, with coefficient 1. The two other $T$ terms have strictly larger $M$, hence strictly smaller complementary leading cuts. Induct on interval length and then downward on $M$. This proves $(U)$. Therefore the $T$ words span precisely the same interval negative coefficient space as the $Y$ words.

If $J$ has consecutive components $[{\beta }_1,{\gamma }_1],\cdots ,[{\beta }_q,{\gamma }_q]$, matrix commutation and $(T)$ give, for $i<s,e_i{T}_{r,s}(J)v=0$ unless $i={\beta }_a-1$ for some $a$; $e_{{\beta }_a-1}{T}_{r,s}(J)v=-T_{r,{\beta }_a-2}(J\cap (r,{\beta }_a-1))T_{{\beta }_a,s}(({\beta }_a,{\gamma }_a]\cup {\bigcup }_{b>a}[{\beta }_b,{\gamma }_b])v$. $(X)$ Empty prefix operators are interpreted as 1. The endpoint action, with $h=\max J$, is $e_{s}T(J)v=T_{r,s-1}(J\setminus \{h\})v+B(h,s)T_{r,s-1}(J)v$ if $h<s$; $e_{s}T(J)v=T_{r,s-1}(J\setminus \{s\})v$ if $h=s$; $e_{s}T(\mathrm{\varnothing })v=B(r,s)T_{r,s-1}(\mathrm{\varnothing })v$. $(E)$

These formulas follow directly as follows. In the empty-$J$ expansion, group $Y$ words by membership of $i$ and $i+1$ in their cuts. The coefficient of each residual word cancels through $-C(r,i+1)+C(r,i)+(a_i+1)=0$ for $i<s$. At the endpoint the two contributions are $C(r,s)$ and $a_s$, whose sum is $B(r,s)$, giving $(E)$. For general $J$, remove $m=\min J$ in $(T)$. For $i<m-1$ the empty-prefix cancellation is unchanged. At $i=m-1$, the first term has zero action; on the product term the empty-prefix endpoint coefficient is $B(r,m-1)+1=C(r,m)$, since the later negative interval increases its adjacent Cartan value by 1. This cancels the displayed denominator and gives the first component case of $(X)$. For later component starts the positive operator commutes with the initial prefix, and applying the induction to both terms in $(T)$ is exactly the same $(T)$ recursion for the shorter first factor in $(X)$. All other internal indices give two zero terms. Applying the endpoint induction to $(T)$ gives $(E)$; in the singleton case its numerator is $B(r,s)T_{r,s-1}(\mathrm{\varnothing })-B(m,s)T_{r,m-1}(\mathrm{\varnothing })T_{m,s-1}(\mathrm{\varnothing })$, and $B(r,s)-B(m,s)=C(r,m)$ yields $(E)$ by $(T)$. The general case cancels in the same way after deleting the maximal cut. All identities were derived integrally before reduction, so no denominator is cancelled in $F$.

Set ${\mathrm{\Xi }}_{r,s}(J)$ by $e_r{e}_{r+1}\cdots e_s{T}_{r,s}(J)v={\mathrm{\Xi }}_{r,s}(J)v$. The endpoint calculation and $(T)$ give $\mathrm{\Xi }(\mathrm{\varnothing })={\prod }_{a=r}^{s}B(r,a),\mathrm{\Xi }(J)=[\mathrm{\Xi }(J\setminus \{m\})-{\mathrm{\Xi }}_{r,m-1}(\mathrm{\varnothing }){\mathrm{\Xi }}_{m,s}(J\setminus \{m\})]\mathrm{/}C(r,m),m=\min J$. $(C)$ Again this is an integer polynomial: it is the actual integral highest coefficient of a matrix-word action, or follows by the same difference-divisibility induction as $\zeta$. The product in the empty case is essential; the historical printed sum in 1995 Definition two point nine is not used.

If $J$ has a bounded zero-$B$ matching $d:J\to [r,s]$, with image $D$, then ${\mathrm{\Xi }}_{r,s}(J)\equiv {\prod }_{a\in [r,s]\setminus D}B(r,a)(\mathrm{m}\mathrm{o}\mathrm{d}p)$. $(M)$ There is a polynomial proof preventing invalid modular cancellation. The stronger identity, for $m=\min J$ (and $m=s+1$ when $J=\mathrm{\varnothing }$), is ${\mathrm{\Xi }}_{r,s}(J)={\prod }_{t\in [r,s]\setminus D}B(r,t)+[{\prod }_{t=r}^{m-1}B(r,t)]{\sum }_{j\in J}B(j,d(j))P_j(C(r,m))$, where each $P_j$ has integer-polynomial coefficients independent of $r$. The empty case is the defining product. Remove the smallest $m$, put $x=C(r,m),b_t=B(m,t),v=d(m)$, and $D^{\mathrm{\prime }}=D\setminus \{v\}$. The two induction formulas for $J\setminus \{m\}$, starting respectively at $r$ and $m$, have exactly the same $P_j$. Their left factors differ by replacing each $b_t$ with $x+b_t$. Substitute them into $(C)$, factoring $A={\prod }_{t=r}^{m-1}B(r,t)$. The principal numerator is ${\prod }_{t\in [m,s]\setminus D^{\mathrm{\prime }}}(x+b_t)-{\prod }_{t\in [m,s]\setminus D^{\mathrm{\prime }}}{b}_t=x{\prod }_{t\in [m,s]\setminus D}(x+b_t)+b_v[{\prod }_{t\in [m,s]\setminus D}(x+b_t)-{\prod }_{t\in [m,s]\setminus D}{b}_t]$. The bracket on the second line is divisible by $x$ in the integer polynomial ring. Every old error term is likewise a difference $R(x)-R(0)$, divided by $x$, because its prefactor and $P_j$ undergo the same translation. Thus after the exact division the first term is the required product with $D$ removed, the second is an integer-polynomial multiple of the new matched generator $b_v=B(m,d(m))$, and the remaining terms are multiples of the old matched generators. For complete control of the universal cancellation, replace the variable $a_r$ by the formal coordinate $x=C(r,m)$, since $m>r$ and its coefficient in $C(r,m)$ is 1. Thus the integer polynomial ring is $\mathbb{Z}[x,\text{remaining }a]$. Every matched generator $B(j,d(j))$ has $j\ge m$ and is independent of $x$. Its quotient is therefore a polynomial ring in $x$ over the remaining coefficient quotient, so multiplication by $x$ is injective even if that coefficient quotient has torsion or zero divisors. The exact paired divisions retain the matched ideal before specialization; no cancellation in a characteristic-$p$ coefficient field occurs. This proves the stronger identity, hence $(M)$, without any division modulo $p$.

$T_{r,s}(J)v=0$ if and only if there is an endpoint $u\in (r,s+1]\setminus J$ and a bounded zero-$B$ matching $\{r\}\cup (J\cap (r,u))\to [r,u)$. $(Z)$ Write $J$ as above and set ${\beta }_{q+1}=s+2$. Every possible witness can be enlarged within its gap to $u={\beta }_a-1$ for some $a\in [1,q+1]$, without changing its matched left set. Thus $(Z)$ is equivalent to one of the following finite conditions $\pi (a)$: the earlier components ${\bigcup }_{b<a}[{\beta }_b,{\gamma }_b]$ have a bounded zero-$B$ matching into $[r,{\beta }_a-2]$, and there is an unused zero-$B$ target for the additional left point $r$.

Induct on interval length. The single-root case is $f_{r}v=0$ iff $a_r=0$ modulo $p$, by $e_r{f}_{r}v=a_{r}v$ and positive detection, exactly $(Z)$. Suppose $\pi (a)$ holds. For each component-start action in $(X)$, if its index $b<a$, the tail has the shorter witness obtained from the existing matching by taking its start ${\beta }_b$ as the extra left point; if $b\ge a$, the prefix has the earlier witness $\pi (a)$. Thus each such product is zero by the induction. All other $e_i$ with $i<s$ annihilate by $(X)$. If $a\le q$, the endpoint witness is $\le {\beta }_q-1$. Both shorter terms in $(E)$ retain it after the last cut ${\gamma }_q$ is deleted, so they are zero by induction. If $a=q+1$, all $J$ has a regular matching $D$ and its unused zero-$B$ target $r$ gives $\mathrm{\Xi }(J)=0$ by $(M)$. Since $e_{i}T(J)v=0$ for $i<s$, any positive simple-word test of the full interval is either zero immediately or is forced to act in the order $s,s-1,\cdots ,r$: a smaller out-of-order first letter commutes past all the already acted larger letters and kills $T(J)v$. Its only possible surviving value is therefore $\mathrm{\Xi }(J)$, which is zero. Positive detection from the modular foundation already proved gives $T(J)v=0$.

Conversely suppose $T(J)v=0$ but no $\pi (a)$ holds. Its earlier prefix in each $(X)$ is nonzero by induction; otherwise the shorter witness would already give one of the forbidden $\pi (a)$. Separated prefix and tail operators in $(X)$ commute, and their product on $v$ is nonzero whenever both factors are nonzero: choose positive tests raising the two separate factors to nonzero multiples of $v$, commute them across the opposite negative factor, and their product raises the product to the nonzero product scalar. Thus each tail in $(X)$ must be zero.

Work backwards through the components. For the last component, its zero tail gives a bounded zero-$B$ matching of all $[{\beta }_q,{\gamma }_q]$ into $[{\beta }_q,s]$. In general the zero tail beginning ${\beta }_b$ gives a matching of components $b,\cdots ,a-1$ into $[{\beta }_b,{\beta }_a-2]$ for some $a>b$. All later components $a,\dots ,q$ already have a regular matching by the backward induction, using targets $\ge {\beta }_a$. These two target ranges are disjoint, so their union matches all components $b,\cdots ,q$. Thus all $J$ is regular. But $T(J)v=0$ implies $\mathrm{\Xi }(J)=0$, and $(M)$ then supplies an unused target $a$ with $B(r,a)=0$. Match $r$ to that target to obtain $\pi (q+1)$, a contradiction. This proves $(Z)$.

It follows immediately that $J$ is nonessential iff $T_{i,s}(J\cap (i,s])v=0$ for some $i\in [r,s]\setminus J$: its witness is precisely the $t=i$ witness in the essential predicate.

If $J$ is nonessential and $T(J)v\mathrm{\ne }0$, choose $i\in (r,s]\setminus J$ with $T_{i,s}(J\cap (i,s])v=0$ by $(Z)$; $i=r$ would have made $T(J)v=0$ already. Put $E=J\cap (r,i),F=J\cap (i,s],e=\max E$ if nonempty. The splitting identity $(P)$ gives $T(J)v=T((E\setminus \{e\})\cup \{i\}\cup F)v+C(e,i)T(J\cup \{i\})v$, with the first term omitted if $E=\mathrm{\varnothing }$ and $e=r$. Both new subsets are strictly larger than $J$. Repeating terminates because there are finitely many subsets; each nonessential vector is expressed through larger essential ones or zero. Combining with $(U)$ and the modular foundation already proved shows that essential $T$ vectors span the root weight space. If there are no essential sets, the result is the zero space; no division is involved in this straightening.

For $I=[b_1,c_1]\cup \cdots \cup [b_H,c_H]$ and $J=[{\beta }_1,{\gamma }_1]\cup \cdots \cup [{\beta }_q,{\gamma }_q]$, write $I\ll J$ when each block of $I$ is a prefix of a block of $J$: $b_a={\beta }_{u_a},c_a\le {\gamma }_{u_a}$. Put ${\mathrm{\Delta }}_0={\bigcup }_{b<u_1}{J}_b,{\mathrm{\Delta }}_a=(J_{u_a}\setminus I_a)\cup {\bigcup }_{u_a<b<u_{a+1}}{J}_b,{\mathrm{\Delta }}_H=(J_{u_H}\setminus I_H)\cup {\bigcup }_{b>u_H}{J}_b$. Call $I$ linked to $J$ if every ${\mathrm{\Delta }}_a$ admits a bounded zero-$B$ matching to targets $\le b_{a+1}-2$, with $b_{H+1}=s+2$. If $H=0$ this means all $J$ is regularly matched to $(r,s]$. Define $\widehat{J}$ as the least linked $I$ in the subset order; it exists since $I=J$ is linked.

Let $h(I,J)$ be the highest coefficient $Q(I)T(J)v=h(I,J)v$. The action formulas give $h(I,J)=0$ unless $I\ll J$, and $h(I,J)=\pm {\mathrm{\Xi }}_{r,b_1-2}({\mathrm{\Delta }}_0){\prod }_{a=1}^{H-1}{\mathrm{\Xi }}_{c_a,b_{a+1}-2}({\mathrm{\Delta }}_a){\mathrm{\Xi }}_{c_H,s}({\mathrm{\Delta }}_H)$ if $I\ll J$. $(H)$ For $H=0$ use $h(\mathrm{\varnothing },J)={\mathrm{\Xi }}_{r,s}(J)$. The signs are the explicit signs from the occurrences of $(X)$, hence $\pm 1$ and irrelevant to zero/nonzero tests.

To verify $(H)$, factor the positive word exactly as $Q(I)=Q_{r,b_1-2}(\mathrm{\varnothing })Q_{b_1,b_2-2}((b_1,c_1])\cdots Q_{b_H,s}((b_H,c_H])e_{b_1-1}{e}_{b_2-1}\cdots e_{b_H-1}$. This is just the canonical $Q$ word reordered by commutations of letters at distance >1. Apply the rightmost displayed gap letters using $(X)$. A start $b_a$ not among the $\beta$'s gives zero immediately. When starts align, each gap action separates the negative factors into intervals with that simple root removed, so they commute with the other interval's positive factors. The sign contributed is $(-1{)}^H$.

The remaining one-block calculation is completely explicit. For $K\subset (b,d]$ and $b\le c\le d,Q_{b,d}((b,c])T_{b,d}(K)v=(-1{)}^{c-b}{\mathrm{\Xi }}_{c,d}(K\cap (c,d])v$ if $(b,c]\subset K$, and is zero otherwise. Indeed $Q=e_c{e}_{c-1}\cdots e_{b+1}{e}_b{e}_{c+1}\cdots e_d$. Move $e_b$ past $e_{c+1},\cdots ,e_d$, which commute with it when $c>b$. If $b+1$ is a cut, $(X)$ gives $e_b{T}_{b,d}(K)v=-T_{b+1,d}(K\setminus \{b+1\})v$; repeat for $b+1,\cdots ,c-1$. Each letter commutes with the still unacted end letters, and each application deletes exactly one initial cut and contributes $-1$. The last remaining word is $e_c{e}_{c+1}\cdots e_d$ on that tail, exactly ${\mathrm{\Xi }}_{c,d}$. If the first missing initial cut is $k+1$, $(X)$ says $e_k$ annihilates the current tail, so the result is zero. The case $c=b$ is the definition of $\mathrm{\Xi }$.

Apply this one-block formula to the separated factors. It forces $c_a\le {\gamma }_{u_a}$; its remaining cut set is exactly ${\mathrm{\Delta }}_a$, while the initial factor has cut set ${\mathrm{\Delta }}_0$. This proves $(H)$, with the explicit sign $(-1{)}^{H+{\sum }_a(c_a-b_a)}$. Empty intervals contribute 1. This proof uses only $(X)$, the definition of $Q$ and commuting separated roots, and no source lemma as a premise.

For linked $I$, apply $(M)$ to every factor in $(H)$. Thus its value is, up to sign, the product of $B(r,d)$ on the unmatched initial target interval and $B(c_a,d)$ on the unmatched later target intervals. This gives the following pivot assertions.

First, if $T(J)v\mathrm{\ne }0$, then $h(\widehat{J},J)\mathrm{\ne }0$. A zero factor in the initial interval would match the extra left point $r$ to that unused target, together with its ${\mathrm{\Delta }}_0$ matching, yielding $(Z)$. A zero factor at endpoint $c_a$ would let one remove that endpoint from $\widehat{J}$ and match its newly released left point to the unused zero target. The old matchings remain disjoint; if that endpoint was a singleton block, merge the two adjacent target intervals, whose old ranges were disjoint. This produces a smaller linked subset, contradicting minimality of $\widehat{J}$.

Second, $h(I,J)=0$ whenever $I>\widehat{J}$. Otherwise $(H)$ gives $I\subset J$. Let $i=\max (I\setminus \widehat{J})$. The two sets agree above $i$. It must be the endpoint of its $I$ block: if $i+1$ were in that same block and in $\widehat{J}$, it would start a $\widehat{J}$ block at $i+1$, hence also a $J$ block by $\ll$, contradicting $i\in J$. The next block start is common to both sets. The link matching for $\widehat{J}$ therefore matches $\{i\}$ together with the corresponding $\mathrm{\Delta }$ block of $I$ into the target interval up to that next start minus 2. Apply $(M)$ to the $\mathrm{\Xi }$ factor starting at $i$. The unused target of $i$ is a zero factor in its residual product, so that factor and $h(I,J)$ are zero.

Third, essential $J$ have distinct $\widehat{J}$. Suppose $J>K$ are essential with the same $\widehat{J}$, and let $t=\max (J\setminus K)$. Their tails above $t$ agree, and $t\mathrm{\notin }\widehat{J}$. Locate $t$ inside the $\mathrm{\Delta }$ block between two consecutive $\widehat{J}$ components. The link matching for that block restricts to a bounded zero-$B$ matching on $\{t\}\cup (K\cap (t,u))$ into $[t,u)$, where $u$ is one less than the next common block start (or $s+1$ at the end). Indeed the $J$ component containing $t$ ends before that next start; all entries between $t$ and $u$ in $K$ are the common tails of $J$. The lower bound follows from boundedness of the matching, and the upper bound is exactly the link target bound. Both $t$ and $u$ lie outside $K$. This makes $K$ nonessential, a contradiction.

In a putative nontrivial relation among essential $T$ vectors, choose $M$ whose $\widehat{J}$ is largest among the nonzero coefficients. Apply $Q(\widehat{M})$. A term with a smaller $\widehat{J}$ is killed by the second pivot assertion, and a term with a larger $\widehat{J}$ has coefficient zero by choice. Distinctness of hats leaves only the $M$ term, with nonzero coefficient $h(\widehat{M},M)$ by the first assertion and essentiality through $(Z)$. This contradicts the relation. Thus essential vectors are independent; the essential straightening already proved supplied spanning. The exact dimension is therefore ${\epsilon }_a(r,s)=\mathrm{\#}\{\text{essential subsets}\}$.

All entries of the $T\mathrm{/}Q$ matrices are reduced integer polynomials in the $a_i$. The construction over $F(t)$ hence gives the same rank and basis over $F$: a relation over $F$ would remain one over $F(t)$, and the equality of finite Gram ranks on extension rules out a missing dimension. Empty root weight spaces and all modular zeros were retained. The proof works for arbitrary dominant polynomial weights and all intervals, including the shifted lower weights needed by first-level character inversion. It does not assert semisimplicity of the Schur or symmetric-group algebra.

Fix a prime $p$, a rank $l\ge 1$, and arbitrary nonnegative fundamental-weight coordinates $\mathrm{\Lambda }=({\lambda }_1,\cdots ,{\lambda }_l)$. No $p$-restricted or $p$-small coordinate assumption is imposed. All $B,C,\lambda$ zero/nonzero tests below are in $F_p$; all cardinalities are ordinary nonnegative integers. For $1\le a\le b\le l$ define $B(a,b)=b-a+{\sum }_{k=a}^b{\lambda }_k$, $C(a,b)=b-a+{\sum }_{k=a}^{b-1}{\lambda }_k$. The identities used throughout are $B(a,b)=C(a,b)+{\lambda }_b$ and, for $a\le x\le b$, $B(a,b)=C(a,x)+B(x,b)$ and $C(a,b)=C(a,x)+C(x,b)$. For integers $r\le s$, $(r,s]=\{r+1,\cdots ,s\}$. An injection $f:M\to N$ is weakly increasing here when $f(j)\ge j$ for every $j$; it need NOT preserve the order of different source elements. Weakly decreasing means $f(j)\le j$. This agrees with Definition one point twelve and is important for the augmenting paths below.

A subset $J\subset (r,s]$ is nonessential if there are $t<u$ in $[r,s+1]\setminus J$ and an injection $d:\{t\}\cup (J\cap (t,u))\to [t,u)$ with $d(x)\ge x$ and $B(x,d(x))=0$ for every source $x$. Otherwise $J$ is essential. Write $E(r,s)$ for this family and $\epsilon (r,s)=\mathrm{\mid }E(r,s)\mathrm{\mid }$. A subset $J\subset (r,s]$ is regular if it has an injection $f:J\to (r,s]$ with $f(j)\ge j$ and $B(j,f(j))=0$. The empty subset is regular.

An index $a$ is normal in $[a,b]$ if ${\lambda }_a\mathrm{\ne }0$ and its demand set $M=\{m\in (a,b]:B(a,m)=0\}$ has a compensation injection $z:M\to (a,b]$ with $z(m)\le m$ and $B(z(m),m)=0$. Define $\sigma (a,b)$ as its indicator. Also set $\sigma (a,a-1)=1$. The root normality depends only on $[a,b]$; an ambient earlier endpoint does not change it.

If $r$ is not normal, put $N(r,s)=\mathrm{\varnothing }$. If $r$ is normal, set $N(r,s)=\{a\in [r,s]:a\text{ is normal in }[a,s]\text{ and }C(r,a)=0\}$, $\nu (r,s)=\mathrm{\mid }N(r,s)\mathrm{\mid }$. The primed functions refer to the last-interval-coordinate shift: ${\epsilon }_{\mathrm{\Lambda }}^{\mathrm{\prime }}(r,s)={\epsilon }_{\mathrm{\Lambda }+{\omega }_s}(r,s)$, ${\nu }_{\mathrm{\Lambda }}^{\mathrm{\prime }}(r,s)={\nu }_{\mathrm{\Lambda }+{\omega }_s}(r,s)$. Only ${\lambda }_s$ is increased by 1. An interval ending before $s$ sees no effect of that shift. All shifted weights remain dominant.

The desired count recurrence, for every $1\le r\le s\le l$, is ${\epsilon }_{\mathrm{\Lambda }}(r,s)={\nu }_{\mathrm{\Lambda }}(r,s)+{\sum }_{u=r+1}^s{\nu }_{\mathrm{\Lambda }}(u,s){\epsilon }_{\mathrm{\Lambda }}^{\mathrm{\prime }}(r,u-1)$. $(R)$ In particular $r=i,s=l$ gives the first-level count formula with the final fundamental coordinate shifted at $u-1$, not at $l$ or at the first coordinate.

Order subsets colexicographically: at the largest element of their symmetric difference, the subset containing that element is larger. The regular family is finite and nonempty, so it has a unique maximum $P$. Claim: an essential regular subset exists iff $r$ is normal; if it exists it is exactly $P$.

First let $J$ be regular and essential, with matching $f$. If ${\lambda }_r=0$, add the source $r$ with $f(r)=r$; this is disjoint from $f(J)$, whose images are $>r$. It is a forbidden witness with endpoints $r,s+1$. Hence ${\lambda }_r\mathrm{\ne }0$. Let $R=\{m\in (r,s]:B(r,m)=0\}$. Every $m\in R$ must belong to $f(J)$, since an unused $m$ could be assigned to the extra source $r$ and would again give a forbidden witness. For each $m\in R$ choose $j=f^{-1}(m)$. Then $j\le m$ and $C(r,j)=B(r,m)-B(j,m)=0$; hence $j$ compensates $m$. These $j$ are distinct, proving $r$ is normal. Moreover, if $I>J$ were another regular subset, let $t$ be the largest element in $I\setminus J$ at the highest difference. The subsets agree above $t$. Restrict a regular matching of $I$ to $\{t\}\cup (J\cap (t,s])$; its targets lie in $[t,s+1)$ and are distinct. Since $t\mathrm{\notin }J$, this is a forbidden witness for $J$. Thus $J=P$.