Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 111 / 3

The Generic Hecke algebra of a Coxeter system is the free module with basis $(T_w{)}_{w\in W}$ over the polynomial ring in parameters $a_i$, subject to $a_i=a_j$ whenever $x_i$ and $x_j$ are conjugate, and with multiplicationEquivalently, its generators satisfy the Coxeter braid relations andIn type $A_n$, all simple generators are conjugate, so there is one parameter.

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A BN-pair consists of subgroups $B,N\le G$ for which $G=⟨B,N⟩$, the subgroup $H=B\cap N$ is normal in $N$, the quotient $W=N/H$ is generated by a distinguished set $S$ of involutions, and the Bruhat multiplication and nondegeneracy axioms hold. The quotient is the associated Weyl group, and the axioms give the Bruhat decomposition of a BN-pair

The Iwahori-Hecke algebra of a BN-pair may be defined, up to the usual opposite-algebra convention, byIts standard basis $(T_w{)}_{w\in W}$ is indexed by the Bruhat double cosets. For a simple generator $x_i$ represented by ${\dot{x}}_i\in N$, setThe double-coset multiplication rule is the generic rule from part a with $a_i$ specialized to $q_i\cdot 1_k$. Thus $H_k(G,B)$ is a specialization of the generic algebra.

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Every Hecke parameter of a BN-pair $q_i$ divides $|B|$ and, because $B$ is a $q$-group, is a power of $q$. If $q\equiv 1(\mathrm{mod}p)$, then $q_i=1$ in the field $k$ of characteristic $p$. The specialized quadratic relation becomeswhile the braid relations are unchanged. These are the defining relations of the Coxeter group $W$, so $x_i\mapsto T_i$ induces a surjective homomorphismBoth algebras have bases indexed by $W$, hence the homomorphism is an isomorphism of group algebras.

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There is a missing hypothesis in the printed claim: it is false when every irreducible component of $W$ has type $A_1$. The intended statement holds as soon as $W$ has an irreducible component of rank at least two, which we now assume.

Since $q=p$, every Hecke parameter of a BN-pair vanishes in $k$, and $H$ is the 0-Hecke algebra withLet $w_0$ be the Longest element of a finite Coxeter group. Choose a simple generator $s$ in a component of rank at least two, putThe element $r$ is again a simple generator. The identities $rv=w_0$ and $\ell (w_0)=\ell (v)+1$ giveIf $t\ne r$ is simple, then $t$ is a left descent of both $v$ and $w_0$: using $\ell (w_{0}u)=\ell (w_0)-\ell (u)$, one gets $\ell (tv)=\ell (v)-1$. HenceThe one-dimensional subspace $kX$ is therefore a left ideal. It is nonzero because $T_v$ and $T_{w_0}$ are distinct basis elements.

Every reduced expression in a Coxeter group for $v=w_{0}s$ contains $r$: in an irreducible finite component of rank at least two, deleting one final generator from $w_0$ does not remove any vertex from its support. A reduced expression for $w_0$ contains $r$ as well. Since $T_{r}X=0$, associativity now givesIf $H$ were a semisimple algebra, the left ideal $kX$ would be a direct summand of the regular module. The corresponding projection would produce a nonzero idempotent in $kX$, impossible because $(kX{)}^2=0$. Thus $H$ is not semisimple.

For completeness, if $W\cong (A_1{)}^m$, thenwhich is semisimple. This is the counterexample showing why the omitted rank condition is necessary.

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