The type-A q-Schur algebra and its Weyl modules

The integral Hom test. The cyclic right module $M^{\nu }$ has presentation $H\mathrm{/}{\sum }_i(T_i-q)H$, where $i$ ranges over the generators of $S_{\nu }$. Indeed, a length-additive decomposition $w=ud$ reduces $T_w$ in this quotient to $q^{\mathrm{\ell }(u)}{T}_d$, and the map $T_d\mapsto m_{\nu }{T}_d$ maps the resulting spanning family bijectively to the independent disjoint-coset basis. Therefore evaluation at $m_{\nu }$ identifies ${\mathrm{Hom}}_H(M^{\nu },M^{\mu })$ with the elements $y\in M^{\mu }$ satisfying $y{T}_i=qy$ for those generators. Pairing $T_w,T_{w{s}_i}$ with $\mathrm{\ell }(w{s}_i)=\mathrm{\ell }(w)+1$ in the standard basis shows that this equation forces their two coefficients to be equal: the coefficient equations are $qb=qa$ and $a+(q-1)b=qb$, so $a=b$ since $q$ is a unit. Requiring all these equalities makes the coefficients constant on each right $S_{\nu }$-coset. Its unweighted coset sum is $T_c{m}_{\nu }$ for the minimal right-coset representative $c$, by length additivity. Hence this simultaneous right eigenspace is $H{m}_{\nu }=(M^{\nu }{)}^{\ast }$, over every coefficient ring. Evaluation consequently gives an isomorphism ${\mathrm{Hom}}_H(M^{\nu },M^{\mu })\cong M^{\mu }\cap H{m}_{\nu }$. For $y=a{m}_{\nu }$ its inverse is $m_{\nu }h\mapsto yh$, which is well defined because $m_{\nu }h=0$ implies $yh=a{m}_{\nu }h=0$.

The semistandard intersection. Expand $y\in H$ uniquely in the Murphy basis. By [F2], membership in $M^{\mu }$ requires zero coefficients outside semistandard first-index fibres and constant coefficients inside each fibre. Apply $\ast$ to [F2] to see that membership in $H{m}_{\nu }$ imposes the identical condition on the second-index fibres of type $\nu$. The two requirements together say precisely that the coefficients are constant on each rectangle $\{(s,t):\mu (s)=S,\nu (t)=T\}$, and zero outside these rectangles. Thus their unweighted sums $m_{ST}$ form a basis of the intersection. The rectangles are disjoint and have a coefficient $1$ at every member, so independence and coefficient extraction hold over arbitrary $R$, including singular parameters; no quantum factorial is cancelled. Every nonempty rectangle has shape with at most $d$ rows, since a strictly increasing column can use only $d$ labels. Combining this with step 1.1 gives exactly the maps ${\phi }_{ST}$ and the asserted basis of each Hom block, hence of $S$. The construction also commutes with arbitrary base change, because the same free basis and formulas describe each specialized Hom block.

The involution under composition. Define $f^{\ast }$ on a Hom block by replacing its evaluation element $y=f(m_{\nu })\in M^{\mu }\cap H{m}_{\nu }$ by $y^{\ast }\in M^{\nu }\cap H{m}_{\mu }$, and extend linearly to $S$. Step 2.1 gives ${\phi }_{ST}^{\ast }={\phi }_{TS}$, so this is an involution. To check reversal, let $f:M^{\nu }\to M^{\mu }$ have $f(m_{\nu })=a{m}_{\nu }$ and $g:M^{\kappa }\to M^{\nu }$ have $g(m_{\kappa })=b{m}_{\kappa }=m_{\nu }h$. Then $(fg)(m_{\kappa })=ab{m}_{\kappa }$, so $(fg{)}^{\ast }(m_{\mu })=m_{\kappa }{b}^{\ast }{a}^{\ast }$. On the other hand $f^{\ast }(m_{\mu })=m_{\nu }{a}^{\ast }$ and $g^{\ast }(m_{\nu })=m_{\kappa }{b}^{\ast }$, whence $(g^{\ast }{f}^{\ast })(m_{\mu })=m_{\kappa }{b}^{\ast }{a}^{\ast }$. Noncomposable blocks give zero on both sides. Since each block map is determined by its generating element, $(fg{)}^{\ast }=g^{\ast }{f}^{\ast }$ as claimed.

The cellular multiplication law. Fix $T$ of shape $\lambda$ and type $\nu$, and first take $f:M^{\kappa }\to M^{\nu }$. Write $f(m_{\kappa })=m_{\nu }h$. Then $({\phi }_{ST}f)(m_{\kappa })=m_{ST}h$. By [F3], its shape-$\lambda$ Murphy coefficient at $(s,v)$ is ${\sum }_{t:\nu (t)=T}{r}_{tv}(h)$ if $\mu (s)=S$, and zero otherwise; all remaining shapes strictly dominate $\lambda$. This coefficient is independent of $s$. The full element lies in $M^{\mu }\cap H{m}_{\kappa }$ by step 1.1, so step 2.1 forces these coefficients to be constant on semistandard second-index fibres $\kappa (v)=V$ and zero on the other fibres. For a fixed $T,h$, these constants are independent of $S$: choose any member $s$ of its nonempty first fibre and read the displayed sum, which contains no $s$. Calling the constants $r_V$, evaluation and the intersection basis give ${\phi }_{ST}f\equiv {\sum }_{V\in T_0(\lambda ,\kappa )}{r}_V{\phi }_{SV}$ modulo the span of strictly dominant Schur shapes, for every $S$. Sum over Hom blocks for arbitrary $f\in S$. Step 3.1 gives the corresponding left law, so each upward span is a two-sided ideal and the cellular axioms hold. This proves cellularity by actual composition and Murphy coefficient comparison.

Partition projections and nonzero forms. For the canonical semistandard tableau $T^{\lambda }$ of shape and type $\lambda$, its refinement fibre consists just of $t^{\lambda }$, so $m_{T^{\lambda }{T}^{\lambda }}=m_{\lambda }$. Accordingly ${\phi }_{T^{\lambda }{T}^{\lambda }}$ is the identity on $M^{\lambda }$ and zero on other summands; it is the genuine projection $1_{\lambda }$. Squaring it gives $1_{\lambda }^2=1_{\lambda }$. In the free shape-$\lambda$ cell quotient, the cell product is $1_{\lambda }^2\equiv B_{\lambda }({\phi }_{T^{\lambda }},{\phi }_{T^{\lambda }})1_{\lambda }$, and the coefficient of this basis vector is therefore $1$. Over a field this form is not identically zero. Step 4.1 and [F4] consequently show that every $W^{\lambda }$ has the nonzero absolutely simple head $L^{\lambda }$, that these heads give all simple modules exactly once, and that the stated diagonal and dominance decomposition conditions hold.

Weight ranks and edge cases. Direct summand projections $1_{\nu }$ satisfy the displayed orthogonality and sum-to-one relations. The source support of ${\phi }_{ST}$ implies ${\phi }_{ST}{1}_{\nu }={\delta }_{\mathrm{type}(T),\nu }{\phi }_{ST}$, so the same formula holds on the right Weyl cell basis, proving its weight rank $K_{\lambda \nu }$. It asserts no idempotence for other cellular diagonal elements. Deleting empty rows leaves the Young subgroup unchanged and identifies the row-type fibres after the order-preserving relabelling of the remaining row labels; hence all constructions allow zero parts. At $n=0$ the unique empty tableau and empty partition give $H=R$, $S=R$ and cell form $1$. If $d=0<n$, the indexing set of weak compositions and the direct sum are empty, $S=0$, and every basis and simple-label assertion is empty. No division except by powers of the unit $q$, generic semisimplicity or choice principle enters any step.