The rank-one quadratic relation in the finite Hecke algebra

For $b\in B$ put $m:=\dot{s}b\dot{s}$; its entries are $m_{kl}=b_{s(k),s(l)}$. For $k>l$, the adjacent transposition satisfies $s(k)>s(l)$ except when $(k,l)=(i+1,i)$; upper triangularity of $b$ therefore gives $m_{kl}=0$ outside that exceptional pair, and $m_{i+1,i}=b_{i,i+1}$. Also $m_{kk}=b_{s(k),s(k)}\mathrm{\ne }0$ by [F4]. If $b_{i,i+1}=0$ then $m$ has all below-diagonal entries zero, so $m\in B$. If $b_{i,i+1}\mathrm{\ne }0$, compute the southwest rank matrix $r_{a,b}(m)$: for $a\le i$ or $a\ge i+2$, columns less than $a$ vanish in rows $a,\dots ,n$, and the square minor on rows and columns $a,\dots ,b$ (when $b\ge a$) has nonzero determinant. If that minor contains both $i,i+1$, it is upper block triangular with central block $\left(\begin{array}{cc}{b}_{i+1,i+1}& 0\\ b_{i,i+1}& b_{ii}\end{array}\right)$ and all other diagonal blocks of size one; otherwise it is upper triangular. In its determinant expansion, the only possible nonidentity permutation would exchange $i,i+1$, whose upper entry is zero. Its determinant is therefore the product of its nonzero diagonal entries, giving $r_{a,b}(m)=\max (0,b-a+1)$. For $a=i+1$, columns below $i$ vanish, columns $i,i+1$ are supported only in row $i+1$, and column $i$ has the nonzero entry $b_{i,i+1}$. Columns $i+2,\dots ,b$, when present, have independent nonzero diagonal entries in rows $i+2,\dots ,b$. Thus the rank is $0$, $1$, $1$ or $b-i$ according as $b\le i-1$, $b=i$, $b=i+1$ or $b\ge i+2$. These numbers equal $\mathrm{\#}\{k\le b:s(k)\ge a\}$ in every case, so by the cell determination of [F3] one has $m\in B\dot{s}B$. Hence $\dot{s}B\dot{s}\subseteq B\cup B\dot{s}B$, and therefore $(B\dot{s}B)(B\dot{s}B)=B(\dot{s}B\dot{s})B\subseteq B(B\cup B\dot{s}B)B=B\cup B\dot{s}B$. Moreover $U_s^{-1}=(B\dot{s}B{)}^{-1}=B{\dot{s}}^{-1}B=B\dot{s}B$ by [F6] and [F5].

Let $\mu :U_s\times U_s\to G$, $\mu (x,y)=xy$, and let $B$ act on the source by $b\cdot (x,y):=(x{b}^{-1},by)$; the action is free, stays in $U_s\times U_s$ by [F3], and $\mu$ is invariant, so every fiber ${\mu }^{-1}(g)$ is a union of free orbits and the integer $m(g):=\mathrm{\mid }{\mu }^{-1}(g)\mathrm{\mid }\mathrm{/}\mathrm{\mid }B\mathrm{\mid }$ is finite, being the number of orbits over $g$. For $b,b^{\mathrm{\prime }}\in B$ the map $(x,y)\mapsto (bx,y{b}^{\mathrm{\prime }})$ is a bijection ${\mu }^{-1}(g)\to {\mu }^{-1}(bg{b}^{\mathrm{\prime }})$, so $m$ is constant on each double coset $BgB$. Over the identity, step 1.1 gives ${\mu }^{-1}(1)=\{(x,x^{-1}):x\in U_s\}$, and the orbit of $(x,x^{-1})$ consists exactly of the pairs $\{(x{b}^{-1},b{x}^{-1}):b\in B\}$, so these orbits correspond to the left cosets $xB$, $x\in U_s$; by [F2] and $\mathrm{\ell }(w_{s_i})=1$ there are $\mathrm{\mid }{U}_s\mathrm{\mid }\mathrm{/}\mathrm{\mid }B\mathrm{\mid }=q$ of them. Hence $m_1:=m(1)=q$.

Since the image of $\mu$ is $U_s\cdot U_s\subseteq B\cup B\dot{s}B$ by step 1.1, the fiber sizes are $m_1=m(1)$ on $B$ and $m_2:=m(\dot{s})$ on $B\dot{s}B$; counting the source gives $q^2\mathrm{\mid }B{\mathrm{\mid }}^2=\mathrm{\mid }{U}_s{\mathrm{\mid }}^2={\sum }_{g\in G}\mathrm{\mid }{\mu }^{-1}(g)\mathrm{\mid }=\mathrm{\mid }B\mathrm{\mid }(m_1\mathrm{\mid }B\mathrm{\mid }+m_2\mathrm{\mid }B\dot{s}B\mathrm{\mid })=\mathrm{\mid }B{\mathrm{\mid }}^2(m_1+q{m}_2)$ by [F2], hence $q^2=q+q{m}_2$ and $m_2=q-1$. Therefore, using [F1] and $\mathrm{\mid }B{\mathrm{\mid }}^{-1}{\sum }_{g\in B}g=e_B$, $\mathrm{\mid }B{\mathrm{\mid }}^{-1}{\sum }_{g\in B\dot{s}B}g=T_s$, $T_s^2=\frac{1}{\mathrm{\mid }B{\mathrm{\mid }}^2}{\sum }_{x,y\in U_s}xy=\frac{1}{\mathrm{\mid }B\mathrm{\mid }}(m_1{\sum }_{g\in B}g+m_2{\sum }_{g\in B\dot{s}B}g)=q{T}_1+(q-1)T_s\mathrm{.}$ Equivalently $(T_s-q)(T_s+1)=0$, so the minimal polynomial of the operator by which $T_s$ acts on any finite-dimensional complex representation divides $(x-q)(x+1)$ and its eigenvalues lie among $q$ and $-1$.

Step 3.1 proves the displayed quadratic relation, equivalently the factored form, and the eigenvalue statement; all data are finite groups and finite sums with the explicit permutation matrix $\dot{s}$, so no choice principle is used.