Affine closest pairs and downward translation from a lower facet

By [F1], transport $C$ to a fundamental alcove by an affine isometry, and fix an interior point $p$. For $z=gy$ outside $C$, reflect $z$ in any defining wall $H$ of $C$ whose interior halfspace it violates. No orbit point of $y$ lies on a wall. This step strictly decreases $\mathrm{\parallel }z-p\mathrm{\parallel }$, since $p$ is strictly on the opposite side of $H$. The orbit $W_{\mathrm{\ell }}y$ is a finite union of translates of the discrete lattice $\mathrm{\ell }Q$, so every bounded ball contains finitely many of its points. The strictly decreasing procedure therefore terminates in $C$, at $y$: the regular orbit representative in $C$ is unique by simple transitivity and disjoint alcove interiors in [F1].

Let $d_H$ be signed distance to $H$, positive on the interior side of $C$. Orthogonal decomposition normal to $H$ gives $\mathrm{\parallel }z-x{\mathrm{\parallel }}^2-\mathrm{\parallel }{r}_{H}z-x{\mathrm{\parallel }}^2=-4d_H(z)d_H(x)$. At each folding step $d_H(z)<0$ and $d_H(x)\ge 0$, so distance to $x$ never increases, strictly decreasing unless $x\in H$. Exactly one defining wall contains $x$, namely the wall of $F$. Thus equality of the initial and terminal distances permits only $s_F$ steps. After one such step the point is on the interior side of that wall, so a second consecutive $s_F$ is impossible. Therefore $\mathrm{\parallel }x-gy\mathrm{\parallel }\ge \mathrm{\parallel }x-y\mathrm{\parallel }$, with equality exactly when $gy=y$ or $gy=s_{F}y$. Conversely both these points attain equality because $s_{F}x=x$.

Apply step 2.1 with $g=u^{-1}v$. In the first equality case take $\tau =u$; in the second take $\tau =u{s}_F$, using $s_{F}x=x$. This proves the asserted common-isometry equality description. The stabilizer of $x$ is $\{1,s_F\}$: only two alcoves have $x$ in their closures, and simple transitivity in [F1] identifies the two possible images of either incident alcove. Consequently the minimizing regular points paired with $gx$ are precisely $gy$ and $g{s}_{F}y$.

Suppose $F$ is lower. Its defining reflection subtracts the positive multiple $((y,\alpha )-k\mathrm{\ell })\alpha$, so $s_{F}y<y$. Track both $y_0^{+}=y$ and $y_0^{-}=s_{F}y$, together with $x_0=x$, through the prescribed downward reflections. At each stage the regular points lie in the two alcoves adjacent to the unique facet containing the current $x_j$. Any reflecting hyperplane missing $x_j$ has both adjacent interiors on the same side as $x_j$: an alcove interior cannot meet a wall, and if its interior lay on the other side from a point in its closure that wall would contain the closure point. A strict downward step uses a hyperplane missing $x_j$. Hence both transported regular points satisfy its same strict downward inequality and are reflected by subtracting positive multiples of its positive finite root. Induction yields $gy\le y$ and $g{s}_{F}y\le s_{F}y<y$, where $g=r_N\cdots r_1$.

For a regular orbit point $y^{\mathrm{\prime }}$ with $\mathrm{\parallel }{x}^{\mathrm{\prime }}-y^{\mathrm{\prime }}\mathrm{\parallel }\le \mathrm{\parallel }x-y\mathrm{\parallel }$, step 3.1 forces equality and $y^{\mathrm{\prime }}=gy$ or $g{s}_{F}y$. Both are at most $y$ by step 4.1. If the points are integral weights and $\mathrm{\ell }$ is an integer, every subtracted positive-root coefficient is an integer by the weight pairings, including the initial facet reflection; this gives the asserted integral root order. The argument uses only finite reflection sequences and finite choices from fixed lists of defining walls, hence no Axiom of Choice.