Leading Schur coefficients and split good-characteristic asymptotic specializations

On a split matrix algebra, a symmetric linear functional is a scalar multiple of ordinary matrix trace: commuting $e_{ii}$ with $e_{ij}$ forces its value on $e_{ij}$ to be zero for $i\mathrm{\ne }j$, and commuting $e_{ij}$ with $e_{ji}$ makes all diagonal values equal. Thus on a split product it is ${\sum }_E{s}_E{\mathrm{Tr}}_E$. Nondegeneracy forces every $s_E\mathrm{\ne }0$, since otherwise an entire block is orthogonal to the algebra. Apply this to $J_{\mathbb{Q}}$ from [F1], and set $f_E=s_E^{-1}\in {\mathbb{Q}}^{\times }$. The same argument gives the generic Hecke Schur scalars in [F5].

For any finite split algebra with symmetrizing trace $\sum f_E^{-1}{\mathrm{Tr}}_E$ and dual bases $b_r,b_r^{\vee }$, its inverse pairing tensor is basis-independent and equals ${\sum }_E{f}_E{\sum }_{i,j}{e}_{ij}^E\otimes e_{ji}^E$. Indeed both tensors represent the identity map under the trace pairing; in the matrix-unit basis the dual of $e_{ij}^E$ is $f_E{e}_{ji}^E$. Applying the matrix-coefficient functions gives ${\sum }_r{\rho }_G(b_r{)}_{ij}{\rho }_E(b_r^{\vee }{)}_{ab}={\delta }_{G,E}{f}_E{\delta }_{i,b}{\delta }_{j,a}$. Summing the appropriate diagonal entries gives ${\sum }_r{\chi }_E(b_r){\chi }_E(b_r^{\vee })=f_E\dim V_E$. These are actual coefficient identities over $\mathbb{Q}$ or $K$, with no character-orthogonality result left as a citation.

Choose a rational basis of $V_E$ and let $L_0$ be its integer span. The sum ${\mathrm{\Lambda }}_E={\sum }_{w\in W}\mathbb{Z}{\rho }_E(t_w)L_0$ is finitely generated, contains $L_0$ since $1_J\in J$, and is stable under every $t_x$ because the multiplication constants of $J$ are integers. It is torsion-free of rank $d_E=\dim V_E$. The integers form a PID (division with remainder makes each nonzero ideal generated by its least positive member), so [F4] makes ${\mathrm{\Lambda }}_E$ finite free. In such a basis all ${\rho }_E(t_w)$ have integer entries and ${\chi }_E^J(t_w)\in \mathbb{Z}$.

Let $z_E={\sum }_w{\chi }_E^J(t_{w^{-1}})t_w\in J$. The coefficient identity in step 2.1 says that its action on $V_G$ is ${\delta }_{E,G}{f}_E\mathrm{id}$. Left multiplication by $z_E$ on the integral basis of $J$ has an integer matrix. The Cayley–Hamilton identity therefore shows that the rational number $f_E$ is an algebraic integer; a rational root of a monic integer polynomial is an integer, by clearing its reduced denominator in that polynomial. Hence $f_E\in \mathbb{Z}\setminus \{0\}$.

Put $a={\nu }_E$. The character of $E=K\otimes V_E$ satisfies ${\chi }_E(D_x)={\sum }_{z,d:a(z)=a(d)=a}{h}_{x,d,z}{n}_d{\chi }_E^J(t_z)$ by [F1]. Each $v^a{h}_{x,d,z}$ is in $\mathbb{Z}[v]$ by the definition of $a(z)$. Taking its constant term gives ${\sum }_{z,d:a(z)=a(d)=a}{\gamma }_{x,d,z^{-1}}{n}_d{\chi }_E^J(t_z)={\chi }_E^J(t_x)$, since the degree-$a$ identity ${\sum }_{d:a(d)=a}{n}_d{t}_d$ acts as the identity on $V_E$ and all other support degrees act by zero. Thus $v^a{\chi }_E(D_x)\in \mathbb{Z}[v]$ and it is congruent to ${\chi }_E^J(t_x)$ modulo $v\mathbb{Z}[v]$.

To see positivity, use the rational positive definite form $⟨x,y⟩=\mu (x{\iota }_J(y))$ on $J_{\mathbb{Q}}$: [F1] makes its basis $t_w$ orthonormal. Trace symmetry implies $⟨ax,y⟩=⟨x,{\iota }_J(a)y⟩$. A rational simple left ideal $U_E\subset J_{\mathbb{Q}}$ exists from its matrix decomposition and is invariant under both $a$ and ${\iota }_J(a)$; restriction of this positive form to $U_E$ is nondegenerate. Thus their matrices on $U_E$ are adjoints and have equal trace. It follows that ${\chi }_E^J(t_{w^{-1}})={\chi }_E^J(t_w)$. Step 2.1 now gives $f_E{d}_E={\sum }_w{\chi }_E^J(t_w{)}^2>0$. The strict inequality holds because ${\chi }_E^J$ is a nonzero linear functional on $J_{\mathbb{Q}}$, so it cannot vanish on the basis $\{t_w\}$; at least one rational summand is nonzero. Therefore $f_E>0$.

In $J_S$ set $q_{ij}^E=f_E^{-1}{\sum }_w{\rho }_E(t_{w^{-1}}{)}_{ji}{t}_w$. Step 2.1 gives ${\rho }_G(q_{ij}^E)={\delta }_{E,G}{e}_{ij}$. These elements lie in $J_S$ by step 2.2 and send every block to its indicated matrix unit. Hence they provide an inverse to $\rho :J_S\to {\prod }_E{\mathrm{End}}_S(S\otimes {\mathrm{\Lambda }}_E)$: their composites are the identity after extension to $\mathbb{Q}$, and both modules are torsion-free, so they are already the identity over $S$. This proves the displayed algebra isomorphism. Since it is defined by inverse coefficient matrices over $S$, it survives every commutative base change out of $S$, including all stated fields.

Let $j$ be the semilinear algebra involution sending $v$ to $v^{-1}$ and $T_s$ to $-T_s$. It is the product of bar with $†$, so $D_x=j(C_x)$ since $C_x$ is bar invariant. The native negative triangular expansion of $C_x$ therefore gives $D_x=(-1{)}^{\mathrm{\ell }(x)}{T}_x+{\sum }_{y<x}{r}_{y,x}{T}_y$ with every $r_{y,x}\in v\mathbb{Z}[v]$. Its finite unitriangular inverse gives $T_x=(-1{)}^{\mathrm{\ell }(x)}{D}_x+v\mathbb{Z}[v]$-linear combinations of lower $D_y$. Applying step 3.2 yields $v^a{\chi }_E(T_x)\equiv (-1{)}^{\mathrm{\ell }(x)}{\chi }_E^J(t_x)(\mathrm{m}\mathrm{o}\mathrm{d}v\mathbb{Z}[v])$.

Each central $1_i$ acts on $V_E$ as either zero or the identity, with the identity occurring only for $i={\nu }_E$. These are exact integral matrix equalities on ${\mathrm{\Lambda }}_E$, so they remain the same after every base change in step 4.2. All special simple dimensions and support degrees are therefore the stated ones.

The integral matrices ${\rho }_E(t_w)$ from step 2.2 define the action of $J_A$ on the explicitly free module $L_{E,A}=A{\otimes }_{\mathbb{Z}}{\mathrm{\Lambda }}_E$. Composing with the integral homomorphism ${\varphi }_A$ of [F1] gives Laurent-integer Hecke action matrices. Its generic fibre is exactly the simple $E$ by [F1]. For any coefficient ring $R$, set $L_{E,R}=R{\otimes }_A{L}_{E,A}$; it remains free with its displayed basis. When the $f_E$ are units, step 4.2 identifies this same module as the standard module of the corresponding full matrix block of $J_R$. Thus the needed generic irreducible models are genuinely free by construction even over a two-dimensional local base.

Apply step 2.1 to the generic Hecke symmetrizing trace, using the exact dual basis of [F3]. It gives $c_E(v)d_E={\sum }_x{\chi }_E(T_x){\chi }_E(T_{x^{-1}})$. Multiply by $v^{2a}$ and insert step 4.3. The two length signs cancel, and the constant coefficient is ${\sum }_x{\chi }_E^J(t_x){\chi }_E^J(t_{x^{-1}})=f_E{d}_E$ by step 2.1. All higher coefficients have strictly positive powers of $v$. Dividing by the nonzero integer $d_E$ in characteristic zero gives $c_E(v)=f_E{v}^{-2a}$ plus strictly higher powers. Since $a={\nu }_E$ and ${\alpha }_E$ is the Schur-element exponent fixed in [F2], comparison of the lowest power gives ${\nu }_E={\alpha }_E$. This proves the exponent and leading coefficient assertions directly from the trace pairing. It also shows the Schur element is a Laurent polynomial over $\mathbb{Q}$, sufficient for its lowest term; no cyclotomic factorization is assumed or needed.

Step 3.2 expresses each ${\chi }_E(D_x)$ as an integral Laurent combination of the integer characters ${\chi }_E^J(t_z)$; step 4.3 does the same for ${\chi }_E(T_x)$, so all these values lie in $A$. The element $z_E^H={\sum }_x{\chi }_E(T_{x^{-1}})T_x$ therefore lies in the finite free $A$-algebra $H$. By the matrix coefficient identity in step 2.1 it acts on the generic block $E$ as $c_E(v)$ and on all other generic blocks as zero. The Cayley–Hamilton polynomial of its integral left multiplication shows that $c_E(v)\in K$ is integral over $A$. By [F6], write it as a coprime fraction $f\mathrm{/}g$ in the UFD $A$. A monic equation for $f\mathrm{/}g$, multiplied by its denominator, implies $g\mid f^m$; unique factorization forces $g$ to be a unit. Thus $c_E(v)\in A$, so its specialization is defined even when it vanishes. The good-characteristic condition used in step 4.2 concerns $f_E$, not this full Laurent polynomial.