Tits deformation for the type-A Hecke algebra

If $N=0$ then $A=0$ and all fibers are the zero algebra, whose trace form is nondegenerate on the zero space, so $A_x\cong A_y$ trivially by [F1]. Assume $N\ge 1$. The structure constants $c_{ij}^k$ are regular on $D(f)$; clearing denominators in the identity expressing the associativity of $A$ is unnecessary, but for any $z\in D(f)$ the fiber $A_z$ is the $\mathbb{C}$-algebra with basis $e_i(z)$ and structure constants $c_{ij}^k(z)$.

Let $G(z)$ be the $N\times N$ matrix with entries $\mathrm{tr}(L_{e_i{e}_j})$, computed over the ring $\mathbb{C}[z,1\mathrm{/}f]$; its determinant $\mathrm{\Delta }\in \mathbb{C}[z,1\mathrm{/}f]$ is a rational function regular on $D(f)$, so $\mathrm{\Delta }=d\mathrm{/}{f}^m$ for some $d\in \mathbb{C}[z]$ and $m\ge 0$. For $z\in D(f)$ the fiber of $G$ at $z$ is the trace-form matrix of $A_z$ in the basis $e_i(z)$, because base change preserves the structure constants and hence the matrices of the operators $L_{e_i{e}_j}$. By [F1], $A_z$ is semisimple exactly when $\det G(z)\mathrm{\ne }0$, that is exactly when $d(z)\mathrm{\ne }0$; thus the semisimple locus is $Y=D(fd)$. Since $x,y$ have semisimple fibers, $x,y\in Y$ and $Y\mathrm{\ne }\mathrm{\varnothing }$; also $d\mathrm{\ne }0$ in this case, and $Y={\mathbb{A}}^1\setminus V(fd)$ is infinite because a nonzero polynomial has finitely many roots.

Put $B=\mathbb{C}[z,1\mathrm{/}(fd)]$ and $A_Y=B{\otimes }_{\mathbb{C}[z,1\mathrm{/}f]}A$, free of rank $N$ over $B$. For $x\in Y$, evaluation $z\mapsto x+t$ defines ${\phi }_x:B\to R=\mathbb{C}[\text{\hspace{0.17em}⁣}[t]\text{\hspace{0.17em}⁣}]$, because $(fd)(x+t)$ has nonzero constant term and is a unit. The algebra $A_x^{\mathrm{f}\mathrm{o}\mathrm{r}\mathrm{m}}=R{\otimes }_{B,{\phi }_x}{A}_Y$ is finite free with reduction $A_x$ modulo $t$. By [F1] and [F2], it is isomorphic to $R{\otimes }_{\mathbb{C}}{A}_x$. Write this isomorphism in the chosen bases as $P\in {\mathrm{GL}}_N(R)$ and put $h=(\det P{)}^{-1}$, $u=(fd)(x+t{)}^{-1}$.

Let $J_x$ be the ideal in $\mathbb{C}[z,(p_{ai}),h,u]$ generated by $h\det P-1$, $u(fd)(z)-1$, and the cleared multiplication equations $(fd)(z{)}^M({\sum }_k{c}_{ij}^k(z)p_{ak}-{\sum }_{b,c}{p}_{bi}{p}_{cj}{c}_{bc}^a(x))=0$ for all $a,i,j$, with $M$ large enough to clear denominators. Let $Z_x=V(J_x)$. Its projection to the $z$ coordinate has image $E_x=\{z\in Y:A_z\cong A_x\}$: the equations force $P$ invertible and multiplicative, hence unital because a surjective multiplicative map sends the identity to the identity; conversely each algebra isomorphism satisfies them with the indicated $h,u$. Chevalley [F3] makes $E_x$ constructible. The formal matrix of step 1.3 satisfies these same polynomial equations at $z=x+t$.

We show that $E_x$ is infinite. It contains $x$, through $P=1$, $h=1$, $u=(fd)(x{)}^{-1}$. Suppose $E_x$ were finite and put $g(z):={\prod }_{a\in E_x}(z-a)$, a nonzero polynomial with the simple root $x$; then $g$ vanishes on $\pi (Z_x)$. By the strong Nullstellensatz [F4], vanishing on $Z_x=V(J_x)$ gives $g\in \sqrt{J_x}$, so some power $g^N$ lies in the defining equation ideal $J_x$. On the other hand step 1.3 supplies the point $(z=x+t,P(t),h(t),u(t))$ of $Z_x$ with coordinates in the $\mathbb{C}$-algebra $R=\mathbb{C}[\text{\hspace{0.17em}⁣}[t]\text{\hspace{0.17em}⁣}]$: the intertwining equations hold because $P$ is an isomorphism, and the two normalizing equations hold by construction. Evaluating the identity $g^N=\sum A_i{\text{generator}}_i$ at this point gives $g(x+t{)}^N=0$ in the power-series ring. But $g(z)=(z-x)q(z)$ with $q(x)\mathrm{\ne }0$, so $g(x+t)=tq(x+t)$ with $q(x+t)$ a unit of $R$, and $t^{N}q(x+t{)}^N\mathrm{\ne }0$: a contradiction. Hence $E_x$ is infinite.

A constructible subset of the line is a finite union of locally closed subsets; a locally closed subset is $U\cap V(J)$ with $U$ open and $V(J)$ closed in ${\mathbb{A}}^1$, and an infinite one among them has $V(J)$ infinite, hence $V(J)={\mathbb{A}}^1$ and the piece contains the nonempty open $U$. Therefore an infinite constructible subset of $Y$ is cofinite in $Y$: its complement lies in the complement of a nonempty open subset of ${\mathbb{A}}^1$, a finite set. By steps 2.2 and this observation $E_x$ is cofinite in $Y$, and applying the same argument with $y$ in place of $x$ makes $E_y$ cofinite in $Y$ as well. Since $Y$ is infinite, $E_x\cap E_y\mathrm{\ne }\mathrm{\varnothing }$; for $z\in E_x\cap E_y$ one has $A_x\cong A_z\cong A_y$, proving the general assertion.

Apply the general assertion to $A=\mathbb{C}[z,1\mathrm{/}z]{\otimes }_{\mathbb{Z}[v^{\pm 1}]}{H}_v(n)$ with $f=z$, $x=1$ and $y=q$: by [F5] this is free of finite rank $n!$ over $\mathbb{C}[z,1\mathrm{/}z]$, and its fibers at $1$ and $q$ are $H_1(S_n)\cong \mathbb{C}[S_n]$ and $H_q(S_n)\cong e_B\mathbb{C}[{\mathrm{GL}}_n({\mathbb{F}}_q)]e_B$, both semisimple by [F6]. Hence $H_q(S_n)\cong \mathbb{C}[S_n]$ as $\mathbb{C}$-algebras. An algebra isomorphism carries the set of simple modules to the set of simple modules and preserves dimensions, so the number and dimensions of the simple modules agree; in particular, by Specht modules classify the complex irreducibles of $S_n$, the simple modules of the finite Hecke algebra are parametrized by partitions of $n$, after choosing an isomorphism.

Steps 1.1 and 1.2 reduce to the semisimple locus and identify it as a principal open, steps 1.3 and 2.1 set up the formal solution and the constructible incidence image, steps 2.2 and 3.1 prove the image cofinite and obtain $A_x\cong A_y$, and step 4.1 applies this to the Hecke family. AC enters only through [F3] and [F4], as recorded in [F7]; the formal-lifting and trace-form arguments are choice-free.