The denominator quotient has only imaginary cone support

For a reduced word $w=s_{i_1}\cdots s_{i_l}$, each prefix is reduced and its next root $s_{i_1}\cdots s_{i_{r-1}}{\alpha }_{i_r}$ is positive by F3. Telescoping therefore gives $\rho -w\rho ={\sum }_{r=1}^l{s}_{i_1}\cdots s_{i_{r-1}}{\alpha }_{i_r}\in Q^{+},\mathrm{ht}(\rho -w\rho )\ge l\mathrm{.}$ There are finitely many words of length at most any fixed integer. Thus only finitely many $w$ contribute at a bounded depth, and $w\rho =\rho$ forces $l=0$. The alternant has top coefficient one, and its normalization $a$ is invertible by F4. Reindexing the orbit sum by left multiplication gives $s_i{A}_{\rho }=-A_{\rho }$ as an array; finite multiplicities justify every coefficient.

We justify division of these skew arrays without presuming a Weyl action on the whole completion. Fix $i$ and write $x=e^{-{\alpha }_i}$ and $y^{\mathbf{b}}={\prod }_{j\mathrm{\ne }i}{e}^{-b_j{\alpha }_j}$ for $\mathbf{b}\ge 0$. Let $m(\mathbf{b})=-{\sum }_{j\mathrm{\ne }i}{a}_{ij}{b}_j\ge 0$, where nonnegativity is the off-diagonal sign axiom for a GCM. By F7, reflection sends $x$ to $x^{-1}$ and $y^{\mathbf{b}}$ to $x^{m(\mathbf{b})}{y}^{\mathbf{b}}$. Put $p=e^{-\rho }D=P$ by F6. The skewness of $D,A_{\rho }$ implies for either $f=p$ or $f=a$ that its fixed transverse coefficient satisfies $f_{\mathbf{b}}(x)=-x^{m(\mathbf{b})+1}{f}_{\mathbf{b}}(x^{-1})\mathrm{.}$ This is an equality of coefficient arrays. Since its original exponents in $x$ are nonnegative, the equality forces them to be at most $m(\mathbf{b})+1$; hence each $f_{\mathbf{b}}$ is a polynomial. Moreover $p_{\mathbf{0}}=1-x$ by F6. Also $a_{\mathbf{0}}=1-x$: in step 1.1, a reduced word whose first letter is not $i$ already contributes the off-axis root ${\alpha }_j$; if its first letter is $i$ and it has a second letter $j$, reducedness gives $j\mathrm{\ne }i$, while F7 gives $s_i{\alpha }_j={\alpha }_j-a_{ij}{\alpha }_i$, whose ${\alpha }_j$-coordinate is one because the simple roots are independent. All associated roots are positive, so later summands cannot cancel that off-axis coordinate. Thus only $1,s_i$ occur on the axis.

Form $C=p\mathrm{/}a$ by F4. We show by induction on ${\sum }_{j\mathrm{\ne }i}{b}_j$ that every $C_{\mathbf{b}}(x)$ is a polynomial satisfying $C_{\mathbf{b}}(x)=x^{m(\mathbf{b})}{C}_{\mathbf{b}}(x^{-1})$. The base is $C_{\mathbf{0}}=1$. At a nonzero transverse index, the product equation gives $(1-x)C_{\mathbf{b}}=R_{\mathbf{b}}:=p_{\mathbf{b}}-{\sum }_{\begin{array}{c}\mathbf{u}+\mathbf{v}=\mathbf{b}\\ \mathbf{u}\mathrm{\ne }\mathbf{0}\end{array}}{a}_{\mathbf{u}}{C}_{\mathbf{v}}\mathrm{.}$ Every $\mathbf{v}$ in the finite sum has smaller total transverse degree. Step 2.1 and the induction hypothesis make $R_{\mathbf{b}}$ a polynomial satisfying $R_{\mathbf{b}}(x)=-x^{m(\mathbf{b})+1}{R}_{\mathbf{b}}(x^{-1})$, since $m$ is additive. At $x=1$ this gives $R_{\mathbf{b}}(1)=0$, so polynomial division yields $R_{\mathbf{b}}=(1-x)q$ with $q\in \mathbb{Z}[x]$. Its expansion equals $C_{\mathbf{b}}$ by uniqueness of inversion in formal power series. Substitution and cancellation of $1-x$ then give $q(x)=x^{m(\mathbf{b})}q(x^{-1})$. This completes the induction, including rank one where only the base transverse index exists.

Step 3.1 proves exactly $s_{i}C=C$ as an array for every $i$, with each transverse slice finite. Repeating over a finite word gives $wC=C$. If the coefficient of $e^{-\beta }$ is nonzero, all coefficients of $e^{-w\beta }$ are the same nonzero value. Since the original support lies in $-Q^{+}$, this implies $w\beta \in Q^{+}$ for every $w$, proving the asserted cone support. It does not say that $\beta$ is a root at all, so makes no identification with F5's imaginary-root set.

Suppose nonconstant support exists and take a nonzero $\beta$ of least positive height in it. If $\beta (h_i)>0$, then $s_i\beta =\beta -\beta (h_i){\alpha }_i$ is a nonzero element of $Q^{+}$ by step 4.1 and invertibility of the reflection. Its coefficient is the same and its height is smaller, a contradiction. Thus every pairing is nonpositive. Least height exists in the positive integers; finitely many lattice points have that height, so choosing one uses no AC.