Signed alcove products and the oriented affine Hecke basis

A backtrack crosses the same wall twice in opposite directions, so its signed product is $N_s{N}_s^{-1}=1$ or $N_s^{-1}{N}_s=1$. To check a braid move of length $m<\mathrm{\infty }$, examine the associated $2m$-sector dihedral residue. Project a point sufficiently far in $D$ to a plane transverse to its common codimension-two face. The $m$ wall lines through the origin separate that plane into the $2m$ sectors of [F1]. On traversing their cycle once, the signs are $m$ consecutive plus signs followed by $m$ consecutive minus signs, up to cyclic rotation: the roots defining the crossed sides run consecutively through a half-circle, and the sign of their pairing with the projected point changes exactly at the two bounding rays of a half-plane. Starting at the beginning of the plus block, the positive product is one alternating word of length $m$ and the negative product is the inverse of the other alternating word. The Hecke braid relation makes their product $1$. Every cyclic rotation also has product $1$, since every factor is invertible. Cutting the cycle into the two gallery arcs of the braid move proves equality of their signed products. This also proves the assertion for a residue translated away from the base alcove.

For two expressions $g=s_1\cdots s_r\omega =t_1\cdots t_k\eta$ in the stated form, the semidirect decomposition in [F1] gives $\omega =\eta$ and $s_1\cdots s_r=t_1\cdots t_k$ in $W_{\mathrm{a}\mathrm{f}\mathrm{f}}$. By [F2], reduce the affine words by braid moves and backtrack deletions and compare the resulting reduced expressions. Step 1.1 preserves their signed products for the fixed $D$; multiplying by the common terminal $U_{\omega }$ proves the asserted independence. The same backtrack and dihedral-cycle argument works with any initial alcove. Common root-kernel directions do not meet a wall and are recorded only in the terminal frame, so introduce no additional gallery relation.

Choose a reduced expression for $g$. Replacing each inverse factor by $N_s-c_s$ gives the term $N_g$ from retaining every generator. Every other term uses fewer than $\mathrm{\ell }(g)$ simple factors, so repeated simple multiplication from [F3] expresses it as a finite sum of $N_h$ with $\mathrm{\ell }(h)<\mathrm{\ell }(g)$. Thus $E_D(g)=N_g+{\sum }_{\mathrm{\ell }(h)<\mathrm{\ell }(g)}{a}_{h,g}{N}_h$, with finitely many nonzero coefficients. Induction on $\mathrm{\ell }(g)$ shows that the $E_D(g)$ span: length-zero elements are exactly $U_{\omega }$, and the formula expresses $N_g$ through $E_D(g)$ and shorter elements.

In a finite relation among the $E_D(g)$, choose the largest length with a nonzero coefficient. At that length the expansion in step 3.1 has distinct leading terms $N_g$, and all its other terms have smaller length. Their coefficients cannot cancel because the $N_g$ are a basis. Thus no nonzero relation exists. This proves the asserted basis, even when $\mathrm{\Omega }$ is infinite, because every expansion and relation considered has finite support. All choices and inductions here concern finite words and finite-dimensional residue planes.