Length-additive products in the finite Hecke algebra

Fix $u,v\in S_n$ with $\mathrm{\ell }(uv)=\mathrm{\ell }(u)+\mathrm{\ell }(v)$ and consider the multiplication map $\mu :B\dot{u}B\times B\dot{v}B\to G$, $\mu (x,y)=xy$, together with the right-and-left action $b\cdot (x,y):=(x{b}^{-1},by)$ of $B$. Each $B\dot{w}B$ is stable under right and under left multiplication by $B$ by [F3], so the action stays inside the source; it is free because $x{b}^{-1}=x$ forces $b=1$; and $\mu$ is invariant because $x{b}^{-1}\cdot by=xy$. Hence $\mu$ factors through the set of $B$-orbits, whose cardinality is $\mathrm{\mid }B\dot{u}B\mathrm{\mid }\mathrm{\mid }B\dot{v}B\mathrm{\mid }\mathrm{/}\mathrm{\mid }B\mathrm{\mid }=\mathrm{\mid }B\mathrm{\mid }{q}^{\mathrm{\ell }(u)+\mathrm{\ell }(v)}=\mathrm{\mid }B\mathrm{\mid }{q}^{\mathrm{\ell }(uv)}=\mathrm{\mid }B\dot{u}\dot{v}B\mathrm{\mid }$ by [F2] and the hypothesis. The product set $(B\dot{u}B)(B\dot{v}B)$ contains $\dot{u}\dot{v}$ and is stable under left and right multiplication by $B$, since $(B\dot{u}B)(B\dot{v}B)=B\dot{u}B\dot{v}B$ and $B\cdot B\dot{u}B\dot{v}B\cdot B=B\dot{u}B\dot{v}B$; being a $B$-bi-invariant subset of $G$, it is a union of Bruhat cells by [F3], so it contains $B\dot{u}\dot{v}B$ and has at least $\mathrm{\mid }B\dot{u}\dot{v}B\mathrm{\mid }$ elements. Since it is the image of $\mu$, whose orbit set has exactly $\mathrm{\mid }B\dot{u}\dot{v}B\mathrm{\mid }$ elements, the image equals $B\dot{u}\dot{v}B$ and the orbit set maps bijectively onto it.

Inversion length is subadditive, $\mathrm{\ell }(\sigma \tau )\le \mathrm{\ell }(\sigma )+\mathrm{\ell }(\tau )$ for all $\sigma ,\tau \in S_n$: if $(i,j)\in \mathrm{Inv}(\sigma \tau )$ with $i<j$ and $\tau (i)>\tau (j)$, then $(i,j)\in \mathrm{Inv}(\tau )$; otherwise $\tau (i)<\tau (j)$ and $\sigma (\tau (i))>(\sigma \tau )(j)=\sigma (\tau (j))$, so $(i,j)\in {\tau }^{-1}(\mathrm{Inv}(\sigma ))$. Hence $\mathrm{Inv}(\sigma \tau )\subseteq \mathrm{Inv}(\tau )\cup {\tau }^{-1}(\mathrm{Inv}(\sigma ))$, and taking cardinalities, with $\mathrm{\mid }{\tau }^{-1}(\mathrm{Inv}(\sigma ))\mathrm{\mid }=\mathrm{\mid }\mathrm{Inv}(\sigma )\mathrm{\mid }$ because $\tau$ is a bijection, gives the inequality. Consequently, if $w=s_{i_1}\cdots s_{i_{\mathrm{\ell }}}$ is a reduced word, meaning $\mathrm{\ell }(w)=\mathrm{\ell }$, then every partial product $p_j:=s_{i_1}\cdots s_{i_j}$ has $\mathrm{\ell }(p_j)=j$: subadditivity gives $\mathrm{\ell }(p_j)\le j$ and $\mathrm{\ell }(p_j^{-1}w)\le \mathrm{\ell }-j$ because each of these is a product of $j$ respectively $\mathrm{\ell }-j$ simple transpositions of length $1$ by [F5], and that $p_j^{-1}w=s_{i_{j+1}}\cdots s_{i_{\mathrm{\ell }}}$ (the inverse partial product cancels the initial letters), so $\mathrm{\ell }=\mathrm{\ell }(w)\le \mathrm{\ell }(p_j)+\mathrm{\ell }(p_j^{-1}w)\le \mathrm{\ell }(p_j)+(\mathrm{\ell }-j)$ forces $\mathrm{\ell }(p_j)\ge j$.

By step 1.1 the fibers of $\mu$ are unions of free $B$-orbits and there are exactly $\mathrm{\mid }B\dot{u}\dot{v}B\mathrm{\mid }$ orbits, one over each element of $B\dot{u}\dot{v}B$; a free orbit has cardinality $\mathrm{\mid }B\mathrm{\mid }$, so every $z\in B\dot{u}\dot{v}B$ has exactly $\mathrm{\mid }B\mathrm{\mid }$ preimages and ${\sum }_{x\in B\dot{u}B,y\in B\dot{v}B}xy=\mathrm{\mid }B\mathrm{\mid }{\sum }_{z\in B\dot{u}\dot{v}B}z$. Therefore, using [F1], $T_u{T}_v=\frac{1}{\mathrm{\mid }B{\mathrm{\mid }}^2}{\sum }_{x,y}xy=\frac{1}{\mathrm{\mid }B\mathrm{\mid }}{\sum }_{z\in B\dot{u}\dot{v}B}z=T_{uv},$ which is the asserted identity.

Taking $v=s_i$ a simple transposition with $\mathrm{\ell }(w{s}_i)=\mathrm{\ell }(w)+1$ gives $T_w{T}_{s_i}=T_{w{s}_i}$ by step 2.1, and iterating along the factors of a reduced word $w=s_{i_1}\cdots s_{i_{\mathrm{\ell }}}$ (each partial product has length $j$ by step 1.2, so the length hypothesis holds at every step) proves $T_w=T_{s_{i_1}}\cdots T_{s_{i_{\mathrm{\ell }}}}$ by induction on $\mathrm{\ell }$. Applying step 2.1 with the ordered pair $(v,u)$ in place of $(u,v)$, whose hypothesis is exactly $\mathrm{\ell }(vu)=\mathrm{\ell }(v)+\mathrm{\ell }(u)$, gives $T_v{T}_u=T_{vu}$, and $\dot{v}\dot{u}=\dot{vu}$ by the multiplicativity in [F4] identifies the cell indexed by $vu$.

Step 2.1 is the asserted identity $T_u{T}_v=T_{uv}$ for length-additive products, and steps 1.2 and 3.1 derive the simple-reflection case, the reduced-word formula and the reversed-order statement; the argument uses only finite sets, the explicit permutation matrices $\dot{w}$ and the fixed idempotent $e_B$, so no choice principle is used.