Integral divided-power tensor actions generate the type-A q-Schur algebra

Integral matrices. Write the iterated coproduct $E_i$ as its sum of single-position raising operators. The summand at position $j$ has coefficient $v^{h_i(w_{>j})}$; the lowering summand has coefficient $v^{-h_i(w_{<j})}$. A position cannot be used twice. For $a$ distinct raising positions, swapping two consecutive choices changes the exponent by $2$, since changing $i+1$ to $i$ increases $h_i$ by $2$. Summing the orders gives $v^{a(a-1)\mathrm{/}2}[a{]}_v!$ times the coefficient using the original suffix counts: explicitly ${\sum }_{\pi \in S_a}{v}^{2\mathrm{inv}(\pi )}=v^{a(a-1)\mathrm{/}2}[a{]}_v!$, as follows by inserting the largest letter, which multiplies the sum by $1+v^2+\cdots +v^{2(a-1)}$. The lowering calculation uses prefixes and the same identity. Therefore, over the generic field and then as integral matrices, $E_i^{(a)}{e}_w={\sum }_{\begin{array}{c}J\subseteq \{j:w_j=i+1\}\\ \mathrm{\mid }J\mathrm{\mid }=a\end{array}}{v}^{a(a-1)\mathrm{/}2+{\sum }_{j\in J}{h}_i(w_{>j})}{e}_{w(J\mapsto i)},F_i^{(a)}{e}_w={\sum }_{\begin{array}{c}J\subseteq \{j:w_j=i\}\\ \mathrm{\mid }J\mathrm{\mid }=a\end{array}}{v}^{a(a-1)\mathrm{/}2-{\sum }_{j\in J}{h}_i(w_{<j})}{e}_{w(J\mapsto i+1)}\mathrm{.}$ The toral generator acts on content $\alpha$ by ${\left[\begin{array}{c}{\alpha }_j\\ t\end{array}\right]}_v$, integral by [F3]. Thus every generating matrix preserves $T$. No factorial is inverted after specialization.

Transport of the Hecke action. For $w_{\mu }=1^{{\mu }_1}\cdots N^{{\mu }_N}$ define ${\theta }_{\mu }(m_{\mu }{T}_d)=v^{\mathrm{\ell }(d)}{e}_{w_{\mu }d}$. Every word of content $\mu$ occurs once, $\mathrm{\ell }(d)=\mathrm{inv}(w_{\mu }d)$, and each coefficient is a unit; hence ${\theta }_{\mu }$ is an integral bijection. Standard Hecke multiplication on distinguished cosets gives an ascending swap with coefficient $1$, a descending swap with coefficient $q$ plus $(q-1)$ times the original coset, and coefficient $q$ when the two letters coincide. Multiplying the basis vectors by $v^{\mathrm{inv}(w)}$ changes the two unequal-letter swap coefficients to $v$, giving exactly the displayed tensor formula. Consequently this is an actual Hecke action, with quadratic and braid relations inherited from $H$, and $\theta ={⨁}_{\mu }{\theta }_{\mu }$ is a right-module isomorphism.

Commutation can be checked locally. Put $j=i+1$. On the four two-letter words $ii,ij,ji,jj$, the nonzero local raising values are $E(ij)=ii$, $E(ji)=vii$, $E(jj)=v^{-1}ij+ji$, and the nonzero lowering values are $F(ii)=v^{-1}ij+ji$, $F(ij)=jj$, $F(ji)=vjj$. The Hecke values are $H(ii)=qii$, $H(ij)=vji$, $H(ji)=vij+(q-1)ji$, $H(jj)=qjj$. Substitution proves $EH=HE$, $FH=HF$: on $ij$ and $ji$ the raising composites give respectively $qii$ and $vqii$, and on $jj$ they give $vij+qji$; lowering gives the transposed three checks. With one active letter and one outside $\{i,j\}$ the suffix or inverse-prefix multiplier is $1$, and the outside letter has the same comparison with $i$ and $j$, so swapping or retaining that letter commutes with its replacement. With no active letter both operators are zero. Terms of the full coproduct acting outside the adjacent pair only see its total $h_i$, unchanged by its Hecke action; diagonal toral operators depend only on content. This proves commutation for $E_i,F_i,K_j$. Over the generic field divided powers commute too; since their integral matrices are given in step 1.1, their commutator entries are zero Laurent polynomials. Toral binomials commute by content. Thus $\rho$ has image in the actual integral commutant.

Orbit coordinates for that commutant. Fix source content $\nu$ and split its sorted word into contiguous column blocks of lengths ${\nu }_j$. For an $N\times N$ nonnegative integer matrix $D=(d_{ij})$ of total $r$, with column sums $\nu$ and row sums $\mu$, let ${\mathcal{W}}_D$ be the words containing exactly $d_{ij}$ letters $i$ in block $j$, and put $z_D={\sum }_{w\in {\mathcal{W}}_D}{v}^{{\mathrm{inv}}_{\nu }(w)}{e}_w,$ where ${\mathrm{inv}}_{\nu }$ counts inversions separately inside those blocks. Within a block, pairing an ascending word and its descending swap in $y{T}_s=qy$ forces the descending coefficient to be $v$ times the ascending coefficient, since $v$ is a unit. Adjacent swaps connect every arrangement to the uniquely blockwise sorted word $w_D$. Hence all simultaneous $q$-eigenvectors for the source row subgroup are exactly the free span of the $z_D$, whose supports are disjoint and whose coefficient at $w_D$ is one. By the cyclic presentation [F2], evaluation at $e_{w_{\nu }}$ identifies ${\mathrm{Hom}}_H(T_{\nu },T_{\mu })$ with this eigenspace. Write ${\xi }_D$ for the map with that evaluation. These maps give an explicit integral basis, independently of any generic rank comparison, over every ring with a chosen invertible $v$.

The toral element $P_{\mu }={\prod }_j\left[\begin{array}{c}{K}_j;0\\ {\mu }_j\end{array}\right]$ acts on content $\alpha$ by ${\prod }_j{\left[\begin{array}{c}{\alpha }_j\\ {\mu }_j\end{array}\right]}_v$. If $\alpha \mathrm{\ne }\mu$ and both sum to $r$, at least one ${\alpha }_j<{\mu }_j$, making that factor identically zero; if $\alpha =\mu$, every factor is one. Thus $\rho (P_{\mu })=1_{\mu }$. For a diagonal matrix $D=\mathrm{diag}(\mu )$ the orbit map ${\xi }_D$ is this projection. These conclusions remain identities at parameters where several $K_j$ eigenvalues coincide.

Define the triangular order, within fixed row and column sums, by all northeast and southwest rectangle sums ${\sigma }_{ab}(D)=\{\begin{array}{ll}{\sum }_{s\le a,t\ge b}{d}_{st}& a<b,\\ {\sum }_{s\ge a,t\le b}{d}_{st}& a>b\mathrm{.}\end{array}$ Write $D^{\mathrm{\prime }}⪯D$ if every such sum is at most the corresponding sum for $D$. This is an actual partial order: differences of northeast rectangle sums recover each strictly upper entry, differences of southwest sums recover each strictly lower entry, and the fixed row sums then recover the diagonal. Put $\delta (D)={\sum }_{i,j}\mathrm{\mid }i-j\mathrm{\mid }{d}_{ij}$ and $s(D)={\sum }_{a\mathrm{\ne }b}{\sigma }_{ab}(D)$. The diagonal-boundary sums give $\delta (D)={\sum }_{a=1}^{N-1}({\sigma }_{a,a+1}(D)+{\sigma }_{a+1,a}(D))$. Therefore $D^{\mathrm{\prime }}\prec D$ implies $\delta (D^{\mathrm{\prime }})\le \delta (D)$ and $s(D^{\mathrm{\prime }})<s(D)$. The pair $(\delta ,s)$ of nonnegative integers supplies a terminating induction even when intermediate matrices have different row sums.

Extreme-column multiplication. Acting by $E_i^{(b)}$ on $z_D$ transfers $b$ letters from row $i+1$ to row $i$, distributed as $t_j$ among the column blocks, with $0\le t_j\le d_{i+1,j}$ and ${\sum }_j{t}_j=b$. Its expansion in step 4.1 has only the matrices $D+{\sum }_j{t}_j(E_{ij}-E_{i+1,j})$. Choosing the rightmost available columns greedily maximizes every suffix ${\sum }_{j\ge k}{t}_j$ and minimizes every prefix ${\sum }_{j\le k}{t}_j$. A transfer across the row boundary $i$ changes a northeast rectangle only when its row cutoff is $i$, by such a suffix; it changes a southwest rectangle only when its lower row cutoff is $i+1$, by the negative of such a prefix. Thus this greedy matrix dominates every other supported matrix in $⪯$. For several $E_i$ executed in increasing $i$, the source row $i+1$ has not yet been changed, and each operation changes rectangle coordinates at its own boundary only. Consequently simultaneous greedy choices still give a single maximal supported matrix. If at each boundary the target row originally contains zero letters in every selected column, its coefficient is a Laurent unit: to obtain the blockwise sorted final word, all newly present $i$'s in those columns must have been raised, giving exactly one predecessor word and one subset. That predecessor is blockwise sorted, with coefficient one in $z_D$, and step 1.1 gives a single Laurent monomial. Earlier operations only depleted this target row, so its zero entries persist. Multiplying these monomials gives the claimed unit pivot, and uniqueness of successive row transfers rules out an additional path to the same final matrix.

The lower version is equally explicit. Acting by $F_i^{(b)}$ transfers letters from row $i$ to row $i+1$. Greedy selection of the leftmost available columns maximizes southwest prefix increments and minimizes northeast suffix decrements, hence gives the maximal matrix. Execute several $F_i$ in decreasing $i$; each source row is then unchanged before its turn, and the rectangle boundaries are again distinct. If all selected target entries are initially zero, the blockwise sorted final word has exactly one predecessor and chosen subset, so its coefficient is a Laurent monomial by the lowering formula of step 1.1. This proves the same maximal-support and unit-pivot assertion without invoking a Gaussian coefficient or cancelling one.

Upper elimination. Suppose $D$ has a strictly upper entry and let $k>0$ be the largest distance $j-i$ of such an entry. Set $b_i=d_{i,i+k}$ (zero when $i+k>N$) and $D_0=D-{\sum }_{i=1}^{N-1}{b}_i{E}_{i,i+k}+{\sum }_{i=1}^{N-1}{b}_i{E}_{i+1,i+k},X=E_{N-1}^{(b_{N-1})}\cdots E_1^{(b_1)}\mathrm{.}$ Zero terms with column index greater than $N$ are omitted. Thus $X$ executes its operations in increasing $i$. The rightmost nonzero column of source row $i+1$ in $D_0$ is at most $i+k$: its original distance-$k$ entry was removed, and the shifted entry from row $i$ is at $i+k$. There are at least $b_i$ letters at that column. Hence the greedy transfer selects exactly $b_i$ letters there and recovers $D$. Target row $i$ in $D_0$ has zero at column $i+k$, since that entire outer entry was removed and its added entry is at $i+k-1$. Step 7.1 therefore gives $X{\xi }_{D_0}=u_D{\xi }_D+{\sum }_{D^{\mathrm{\prime }}\prec D}{a}_{D^{\mathrm{\prime }}}{\xi }_{D^{\mathrm{\prime }}},u_D\in A^{\times },a_{D^{\mathrm{\prime }}}\in A\mathrm{.}$ Its source is still $\nu$, and its target row sum is the row sum of $D$, so these lower terms belong to that same Hom block. Also $\delta (D_0)=\delta (D)-{\sum }_i{b}_i<\delta (D)$. Every coefficient is explicitly obtained by the finite subset sums in step 1.1 and reading the blockwise sorted word coefficient of step 4.1.

Lower elimination. If $D$ is lower triangular but not diagonal, let $k>0$ be its largest lower distance and put $b_i=d_{i+1,i+1-k}$. Set $D_0=D-{\sum }_{i=1}^{N-1}{b}_i{E}_{i+1,i+1-k}+{\sum }_{i=1}^{N-1}{b}_i{E}_{i,i+1-k},X=F_1^{(b_1)}\cdots F_{N-1}^{(b_{N-1})},$ with zero terms omitted when the column index is out of range. Now operations execute in decreasing $i$. The leftmost source column in row $i$ of $D_0$ is at least $i+1-k$, with at least $b_i$ letters there; the target entry at that column is zero. Therefore step 8.1 gives the same expansion with unique unit pivot at $D$, and again $\delta (D_0)=\delta (D)-{\sum }_i{b}_i$.

Exact recursive preimages and surjectivity. Induct on $(\delta (D),s(D))$. For a diagonal matrix use $R_D=P_{\mu }$ of step 5.1. In either non-diagonal case, $D_0$ has smaller first coordinate, while each $D^{\mathrm{\prime }}\prec D$ in steps 9.1–10.1 has a strictly smaller pair by step 6.1. With their already constructed preimages define $R_D=u_D^{-1}(X{R}_{D_0}-{\sum }_{D^{\mathrm{\prime }}\prec D}{a}_{D^{\mathrm{\prime }}}{R}_{D^{\mathrm{\prime }}})\mathrm{.}$ This is a finite $A$-linear combination of ordered words in simple divided powers and toral weight projections, with completely specified coefficient extraction; $u_D^{-1}$ is only a Laurent monomial. Its image is exactly ${\xi }_D$. The finite matrix inventory and the decreasing nonnegative pair guarantee termination. Since the ${\xi }_D$ are the actual integral Hom basis, their preimages prove surjectivity. Tensoring this expression with any $A$-algebra gives the same basis and preimages; step 4.1 proves that the specialized commutant itself has this basis. Thus specialization commutes with the endomorphism construction, including nonflat specialization and roots of unity.

Identify the adjoint involution. In the coset basis equip $M^{\mu }$ with diagonal form $⟨m_{\mu }{T}_d,m_{\mu }{T}_e⟩={\delta }_{de}{q}^{\mathrm{\ell }(d)}$. Under ${\theta }_{\mu }$ this becomes ordinary tensor dot product. To see that the involution of [F2] is its adjoint, work temporarily over $\mathbb{Q}(v)$ with the standard Hecke trace $t(T_w)={\delta }_{w,1}$, for which $t(T_x^{\ast }{T}_y)={\delta }_{xy}{q}^{\mathrm{\ell }(x)}$ by standard-basis multiplication. Write $p_{\mu }={\sum }_{w\in S_{\mu }}{q}^{\mathrm{\ell }(w)}$, so $m_{\mu }^2=p_{\mu }{m}_{\mu }$; the coset form is $t(x^{\ast }y)\mathrm{/}{p}_{\mu }$ on $M^{\mu }$. If $f(m_{\nu })=y\in M^{\mu }\cap H{m}_{\nu }$, then $f^{\ast }(m_{\mu })=y^{\ast }$ by the actual definition. For $x=m_{\nu }h$ and $z=m_{\mu }k$, both adjoint pairings are $t(k^{\ast }yh)$: use $m_{\mu }y=p_{\mu }y$ in the first, $y{m}_{\nu }=p_{\nu }y$ in the second, and the trace property. Thus $\ast$ is the dot adjoint. The conclusion concerns equality of integral matrices, so it holds over $A$ and every specialization; the temporary nonzero $p_{\mu }$ is never divided in those rings.

The assignment $\tau (E_i)=F_i$, $\tau (F_i)=E_i$, $\tau (K_j)=K_j$ reverses and preserves the generic defining relations: reversing the toral relation replaces its root exponent by the negative one; the mixed commutator and its right side are both fixed; the Serre polynomial is carried to the reversed negative polynomial. It fixes the toral binomials and exchanges divided powers, hence preserves $U_A$. Let $\beta =\mu +a({ϵ}_i-{ϵ}_{i+1})$. For a chosen raising subset, subtracting the reverse lowering exponent in step 1.1 gives $a({\mu }_i-{\mu }_{i+1})+a^2$, whereas $c_{\beta }\mathrm{/}{c}_{\mu }=v^{-a({\mu }_i-{\mu }_{i+1})-a^2}$. Thus the weighted dot form $⟨x,y{⟩}_C=⟨x,Cy⟩$ makes the raising and lowering matrices adjoint; all toral matrices are diagonal and self-adjoint. Its adjoint operator is $C^{-1}{s}^{\ast }C$. Equality on all integral generators, followed by reversal of products and linearity, proves $\rho (\tau (u))=C^{-1}\rho (u{)}^{\ast }C$ for every $u$.

Square-root-free matrices. In step 2.1 the permutation-module word basis is $b_w=v^{\mathrm{inv}(w)}{e}_w$. If raising a subset $J$ of $a$ positions changes $w$ to $w^{\mathrm{\prime }}$, counting inversions involving a changed letter gives $\mathrm{inv}(w)-\mathrm{inv}(w^{\mathrm{\prime }})={\sum }_{j\in J}(\mathrm{\#}i(w_{>j})-\mathrm{\#}(i+1)(w_{<j}))+\left(\genfrac{}{}{0px}{}{a}{2}\right)\mathrm{.}$ The final correction counts pairs of changed positions, counted incorrectly by changing each letter in isolation. Add the exponent in step 1.1 and use $\mathrm{\#}(i+1)(w_{<j})+\mathrm{\#}(i+1)(w_{>j})={\mu }_{i+1}-1$. The coefficient in the $b$ basis is $v^{a(a-{\mu }_{i+1})+2{\sum }_{j\in J}\mathrm{\#}i(w_{>j})}$. Scaling by $v^{a({\mu }_{i+1}-a)}$ gives the displayed $q$-coefficient. For lowering, the analogous inversion difference is ${\sum }_{j\in J}(\mathrm{\#}(i+1)(w_{<j})-\mathrm{\#}i(w_{>j}))+\left(\genfrac{}{}{0px}{}{a}{2}\right)$, giving coefficient $v^{a(a-{\mu }_i)+2{\sum }_{j\in J}\mathrm{\#}(i+1)(w_{<j})}$ and the asserted normalization. These define matrices over $B$. Their Hecke commutation follows from step 3.1 after the injective substitution $B\hookrightarrow A$, so all their polynomial commutator entries are zero already over $B$.

Over $B$ the permutation-basis Hecke formulas force equal coefficients for ascending and descending arrangements within each source block: an ascending word goes to its swap, a descending word to $q$ times its swap plus $(q-1)$ times itself, and the equation $y{T}_s=qy$ uses only the unit $q$. Thus the unweighted sums ${\sum }_{w\in {\mathcal{W}}_D}{b}_w$ give the same orbit Hom basis, denoted ${\mathrm{\Phi }}_D$. The cross-block inversion count $d(D)={\sum }_{j<k,a>b}{d}_{aj}{d}_{bk}$ is constant on ${\mathcal{W}}_D$, so ${\mathrm{\Phi }}_D=v^{d(D)}{\xi }_D$ under extension to $A$. Repeat steps 6.1–11.1 directly with the normalized operators of step 14.1. Their supports and greedy maximal matrices are identical, and every unique pivot is now a single power of $q$, because every contributing subset coefficient is a power of $q$. The recursion therefore uses only $B$-coefficients and $q$-unit pivots and proves generation over $B$. The same orbit presentation and recursion work after every map $B\to R$. Finally, adjoining $z$ with $z^2=q$ gives $R^{\mathrm{\prime }}=R[z]\mathrm{/}(z^2-q)$, free of rank two over $R$, with $z^{-1}=q^{-1}z$; it is faithfully flat even in characteristic two. Quantum assertions requiring a square root may be used there, while any specified map, equality, kernel, cokernel, or exact sequence already defined over $R$ descends by faithful flatness. The square-root-free generation proved here itself needs no such extension. At $r=0$ every construction is the identity map of $A$ or $B$, and at $N=1$ all matrices are diagonal, so the base case covers both edge cases. No Choice principle, generic rank argument, or nonunit denominator is used for generation.