The finite Hecke algebra as a convolution corner and its endomorphism interpretation

For $b\in B$ the right multiplication $x\mapsto xb$ permutes the basis elements of $\mathbb{C}[G]$, so $e_{B}b=\mathrm{\mid }B{\mathrm{\mid }}^{-1}{\sum }_{b^{\mathrm{\prime }}}{b}^{\mathrm{\prime }}b=\mathrm{\mid }B{\mathrm{\mid }}^{-1}{\sum }_{b^{\mathrm{\prime }}}{b}^{\mathrm{\prime }}=e_B$, and likewise $b{e}_B=e_B$; therefore $e_B^2=\mathrm{\mid }B{\mathrm{\mid }}^{-1}{\sum }_{b}b{e}_B=e_B$, so $e_B$ is an idempotent fixed by left and right multiplication by elements of $B$.

The map $\mathbb{C}[G]\to \mathbb{C}[G]e_B$, $g\mapsto g{e}_B$, is $\mathbb{C}[G]$-linear and surjective, and it is constant on the right $B$-orbits by step 1.1, so it factors through $\mathbb{C}[G]{\otimes }_{\mathbb{C}[B]}\mathbb{C}\cong \mathbb{C}[G\mathrm{/}B]$; the resulting map sends the basis element $gB$ to $g{e}_B$, so it is an isomorphism of left $\mathbb{C}[G]$-modules $\mathbb{C}[G]e_B\cong \mathbb{C}[G\mathrm{/}B]$. Composing with the spherical identification of [F2] gives $\mathbb{C}[G]e_B\cong I(1)$, which is clause (1).

The double cosets $B{P}_{\sigma }B$ partition $G$ by [F3], so $\mathbb{C}[G]$ is the direct sum over $S_n$ of the subspaces $\mathbb{C}[B{P}_{\sigma }B]$, and $H=e_B\mathbb{C}[G]e_B$ is spanned by the elements $e_{B}x{e}_B$ with $x\in G$; since $e_B(bx{b}^{\mathrm{\prime }})e_B=(e_{B}b)x(b^{\mathrm{\prime }}{e}_B)=e_{B}x{e}_B$ for $b,b^{\mathrm{\prime }}\in B$ by step 1.1, each double coset contributes the single vector $e_B\dot{\sigma }{e}_B$ for its permutation representative. That vector is nonzero: the coefficient of $\dot{\sigma }$ in $e_B\dot{\sigma }{e}_B=\mathrm{\mid }B{\mathrm{\mid }}^{-2}{\sum }_{b,b^{\mathrm{\prime }}\in B}b\dot{\sigma }{b}^{\mathrm{\prime }}$ is $\mathrm{\mid }B{\mathrm{\mid }}^{-2}\cdot \mathrm{\mid }B\cap \dot{\sigma }B{\dot{\sigma }}^{-1}\mathrm{\mid }\ge \mathrm{\mid }B{\mathrm{\mid }}^{-2}>0$, since $b\dot{\sigma }{b}^{\mathrm{\prime }}=\dot{\sigma }$ exactly when $b=\dot{\sigma }{b}^{\mathrm{\prime }-1}{\dot{\sigma }}^{-1}\in B$, and $b=b^{\mathrm{\prime }}=1$ contributes. Therefore the $n!$ elements $e_B\dot{\sigma }{e}_B$, one per double coset, form a basis of $H$, and ${\dim }_{\mathbb{C}}H=\mathrm{\mid }W\mathrm{\mid }=n!$: this is the dimension assertion of clause (3).

Let $f:\mathbb{C}[G]e_B\to \mathbb{C}[G]e_B$ be $\mathbb{C}[G]$-linear and put $a:=f(e_B)$. Since $e_B$ acts on $\mathbb{C}[G]e_B$ by left multiplication and $f$ is linear over $\mathbb{C}[G]$, one gets $f(g{e}_B)=ga$ for all $g\in G$; also $e_{B}a=a$ and $a{e}_B=a$ because $a=f(e_B)=f(e_B{e}_B)=e_{B}a$ and $a=f(e_B)\in \mathbb{C}[G]e_B$. Hence $a\in H$ and $f(g{e}_B)=(g{e}_B)a$, so $f$ is the right multiplication ${\rho }_a$. Conversely, for $a\in H$ the map ${\rho }_a(y{e}_B)=y{e}_{B}a$ takes values in $\mathbb{C}[G]e_{B}a\subseteq \mathbb{C}[G]e_B$ and commutes with left multiplication by $G$, so it is a $\mathbb{C}[G]$-linear endomorphism, and it satisfies ${\rho }_a{\rho }_b={\rho }_{ba}$ by associativity. The assignment is therefore a $\mathbb{C}$-linear bijection $H\to {\mathrm{End}}_{\mathbb{C}[G]}(\mathbb{C}[G]e_B)$ that reverses composition, i.e. a $\mathbb{C}$-algebra isomorphism onto the opposite algebra; the identification with ${\mathrm{End}}_G(\mathbb{C}[G\mathrm{/}B])$ is transport along the isomorphism of step 2.1. This is the first part of clause (2).

The $\mathbb{C}$-linear map ${\sum }_g{c}_{g}g\mapsto {\sum }_g{c}_g{g}^{-1}$ is an anti-automorphism of the algebra $\mathbb{C}[G]$ because $(gh{)}^{-1}=h^{-1}{g}^{-1}$, it fixes $e_B$ because inversion permutes $B$, and it therefore restricts to an algebra anti-automorphism of $H=e_B\mathbb{C}[G]e_B$; explicitly it sends $e_B\dot{w}{e}_B$ to $e_B{\dot{w}}^{-1}{e}_B$. Hence $H\cong H^{\mathrm{o}\mathrm{p}}$, so the opposite algebra in step 3.1 is isomorphic to $H$ itself and clause (2) is complete.

Finally $H\cong {\mathrm{End}}_G(\mathbb{C}[G\mathrm{/}B])$ by steps 3.1 and 4.1, and $\mathbb{C}[G\mathrm{/}B]$ is a finite-dimensional semisimple $\mathbb{C}[G]$-module by [F4]; hence ${\mathrm{End}}_G(\mathbb{C}[G\mathrm{/}B])$ is a semisimple $\mathbb{C}$-algebra, a product of complex matrix algebras, by [F5], and so is its opposite, which is $H$. This proves the semisimplicity statement of clause (3); clauses (1)-(3) are now established, and no step selected a basis of $\mathbb{C}[G\mathrm{/}B]$ or of any quotient of $G$, so no choice principle is used.