Cartan-number products, allowed edge labels, tree scalings and reflection stability

For all $s,t\in S$ one has $a_{st}=2\frac{c_s}{c_t}B(e_s,e_t)$; in particular $a_{ss}=2$ and $a_s^{\vee }=\frac{2e_s}{c_s}$.

For distinct $s,t$ with $m(s,t)<\mathrm{\infty }$ one has $a_{st}=-2\frac{c_s}{c_t}\cos \frac{\pi }{m(s,t)}\le 0$ and $a_{st}{a}_{ts}=4{\cos }^2\frac{\pi }{m(s,t)}$; for $m(s,t)=\mathrm{\infty }$ one has $a_{st}=-2c_s\mathrm{/}{c}_t<0$ and $a_{st}{a}_{ts}=4$; and $a_{st}=0$ exactly when $m(s,t)=2$. Indeed $2\le m(s,t)<\mathrm{\infty }$ gives $\pi \mathrm{/}m(s,t)\in (0,\pi \mathrm{/}2]$, where $\cos \ge 0$ and $\cos =0$ only at $\pi \mathrm{/}2$ because $\cos$ is strictly decreasing on $[0,\pi ]$ with $\cos (\pi \mathrm{/}2)=0$.

The values $\cos \frac{\pi }{2}=0$, $\cos \frac{\pi }{3}=\frac{1}{2}$, ${\cos }^2\frac{\pi }{4}=\frac{1}{2}$ and ${\cos }^2\frac{\pi }{6}=\frac{3}{4}$ hold, and $\cos x>0$ for $0<x<\frac{\pi }{2}$. For the first, $\cos (\pi \mathrm{/}2)=0$; putting $c:=\cos (\pi \mathrm{/}3)$, the supplementary identity at $x=\pi \mathrm{/}3$ gives $\cos (2\pi \mathrm{/}3)=-c$ while the double-angle identity gives $\cos (2\pi \mathrm{/}3)=2c^2-1$, so $2c^2+c-1=(2c-1)(c+1)=0$ and $c>0$ (as $0<\pi \mathrm{/}3<\pi \mathrm{/}2$ and $\cos$ decreases from $\cos (\pi \mathrm{/}2)=0$) force $c=\frac{1}{2}$; the double-angle identity at $x=\pi \mathrm{/}4$ gives $2{\cos }^2(\pi \mathrm{/}4)-1=\cos (\pi \mathrm{/}2)=0$, and at $x=\pi \mathrm{/}6$ it gives $2{\cos }^2(\pi \mathrm{/}6)-1=\cos (\pi \mathrm{/}3)=\frac{1}{2}$.

If $B$ is positive definite then $(V,B)$ is a real inner product space, and for linearly independent $u,v\in V$ one has $B(u,v{)}^2<B(u,u)B(v,v)$; moreover for distinct $s,t$ the vectors $a_s=c_s{e}_s$ and $a_t=c_t{e}_t$ are linearly independent.

Let $\mathrm{\Gamma }$ be a forest whose edge labels lie in $\{3,4,6\}$, with a root chosen in each component. Then each component is a tree, every vertex $t$ other than its root has a unique neighbour $s$ on its path to that root, and the prescription $c_{s_0}:=1$, $c_t:=2c_s\cos (\pi \mathrm{/}m(s,t))$ determines a unique positive value $c_t$ for every vertex.

Writing $r_s:=\rho (s)=r_{e_s}$ for $s\in S$, the reflection formula gives, for all $s,t$, $r_s(a_t)=a_t-a_{ts}{a}_s$ and $r_s(a_t^{\vee })=a_t^{\vee }-a_{st}{a}_s^{\vee }$.

Assume $B$ positive definite and $c$ crystallographic. Then for distinct $s,t$ one has $0\le a_{st}{a}_{ts}<4$, so $a_{st}{a}_{ts}\in \{0,1,2,3\}$; and $m(s,t)\in \{2,3,4,6\}$, with $4{\cos }^2(\pi \mathrm{/}m(s,t))=0,1,2,3$ for $m=2,3,4,6$ respectively. The bound $<$ uses strict Cauchy-Schwarz in the basis-independent pair $a_s,a_t$; the four values use step 1.3; and no other $m$ occurs because $m=5$ gives $1\mathrm{/}2<{\cos }^2(\pi \mathrm{/}5)<3\mathrm{/}4$ (from $\pi \mathrm{/}6<\pi \mathrm{/}5<\pi \mathrm{/}4$ by decrease of $\cos$), so $4{\cos }^2(\pi \mathrm{/}5)\in (2,3)$, while $7\le m<\mathrm{\infty }$ gives $3\mathrm{/}4<{\cos }^2(\pi \mathrm{/}m)<1$, so $4{\cos }^2(\pi \mathrm{/}m)\in (3,4)$, and $m=\mathrm{\infty }$ gives the product $4$.

Let $\mathrm{\Gamma }$ be a forest with edge labels in $\{3,4,6\}$ and let $c$ be the tree scaling of step 1.5. Then $c$ is crystallographic: on every edge $\{s,t\}$ with $s$ the root-side endpoint, $a_{st}=-1$ and $a_{ts}=-4{\cos }^2(\pi \mathrm{/}m(s,t))\in \{-1,-2,-3\}$; for non-adjacent distinct $s,t$ one has $a_{st}=0$; and $a_{ss}=2$.

Assume $c$ crystallographic. In the formulas of step 1.6, the coefficients $a_{ts}$ and $a_{st}$ are integers by [F11], so each generator matrix $r_s$ has integer entries in both bases $(a_t)$ and $(a_t^{\vee })$. Every $w\in W$ is a finite product of elements of $S$ [F2], and $\rho$ is a homomorphism with $\rho (s)=r_s$ [F6]; therefore the matrices of $\rho (w)$ in both bases have integer entries. In particular, for every $w\in W$ and $t\in S$, $\rho (w)a_t$ is an integral linear combination of the $a_s$. The $a$-basis coefficients of $\rho (w)a_s$ all have one sign because $\rho (w)a_s=c_s\rho (w)e_s$ has, in the $e$-basis, coefficients of one sign by [F9] and [F7], and re-expressing in the $a$-basis multiplies the $t$-th coefficient by the positive factor $c_s\mathrm{/}{c}_t$.

Assume $c$ crystallographic. Step 1.6 and [F11] give $r_s(a_t)\in Q$ and $r_s(a_t^{\vee })\in Q^{\vee }$ for all $s,t$, hence $r_s(Q)\subseteq Q$ and $r_s(Q^{\vee })\subseteq Q^{\vee }$; since $r_s^2=\mathrm{i}\mathrm{d}$ by [F5], applying $r_s$ gives the reverse inclusions, so both are equalities. Since $W$ is generated by $S$ and $\rho (s)=r_s$ [F2, F6], every $\rho (w)$ preserves $Q$ and $Q^{\vee }$. Because $(a_s)$ and $(a_s^{\vee })$ are bases of $V$, their $\mathbb{Z}$-spans $Q$ and $Q^{\vee }$ are free abelian groups of rank $\mathrm{\mid }S\mathrm{\mid }$ and are $\rho (W)$-stable.

Assume $B$ positive definite and $c$ crystallographic, and let $\{s,t\}$ be an edge of $\mathrm{\Gamma }$ with label $m\in \{3,4,6\}$; put $p:=4{\cos }^2(\pi \mathrm{/}m)\in \{1,2,3\}$. Then $a_{st}$ and $a_{ts}$ are negative integers with product $p$, so $\{\mathrm{\mid }{a}_{st}\mathrm{\mid },\mathrm{\mid }{a}_{ts}\mathrm{\mid }\}=\{1,p\}$ and $\frac{c_s^2}{c_t^2}=\frac{a_{st}^2}{p}\in \{1\mathrm{/}p,p\}$; in particular every edge of label $3$ has $c_s=c_t$.

Assume $c$ crystallographic. By step 2.3, ${\mathrm{\Phi }}_c\subseteq Q$, hence $\mathbb{Z}{\mathrm{\Phi }}_c\subseteq Q$, and since each $a_s\in {\mathrm{\Phi }}_c$ also $Q\subseteq \mathbb{Z}{\mathrm{\Phi }}_c$, so $Q=\mathbb{Z}{\mathrm{\Phi }}_c$. Likewise, for $\beta =\rho (w)a_s$ the identity ${\beta }^{\vee }=\frac{2\beta }{B(\beta ,\beta )}=\rho (w)\frac{2a_s}{B(a_s,a_s)}=\rho (w)a_s^{\vee }$ (using preservation of $B$) shows that each element of ${\mathrm{\Phi }}_c^{\vee }$ lies in $Q^{\vee }$ by step 2.4, so $\mathbb{Z}{\mathrm{\Phi }}_c^{\vee }\subseteq Q^{\vee }$, while $a_s^{\vee }=\frac{2a_s}{B(a_s,a_s)}$ with $a_s\in {\mathrm{\Phi }}_c$ gives the reverse inclusion; hence $Q^{\vee }=\mathbb{Z}{\mathrm{\Phi }}_c^{\vee }$. Finally $Q\subseteq P$, since $B(a_s,q^{\vee })={\sum }_t{n}_t{a}_{st}\in \mathbb{Z}$ for $q^{\vee }={\sum }_t{n}_t{a}_t^{\vee }\in Q^{\vee }$ by bilinearity and integrality of the $a_{st}$, so each $a_s$ lies in $P$ and $P$ is an additive subgroup.

Assume $c$ crystallographic. For $\beta =\rho (w)a_s$ and $\gamma =\rho (v)a_t$ in ${\mathrm{\Phi }}_c$ one has $\frac{2\gamma }{B(\gamma ,\gamma )}=\rho (v)a_t^{\vee }$ and $B(\beta ,\rho (v)a_t^{\vee })=B(\rho (v{)}^{-1}\beta ,a_t^{\vee })={\sum }_u{m}_u{a}_{ut}\in \mathbb{Z}$, where $\rho (v{)}^{-1}\beta =\rho (v^{-1}w)a_s={\sum }_u{m}_u{a}_u$ has integer coefficients $m_u$ by step 2.3.

Assume $B$ positive definite, $c$ crystallographic and $\mathrm{\Gamma }$ connected. Then $\mathrm{\Gamma }$ has at most one edge of label $\ge 4$, every other edge has label $3$ and hence squared length ratio $1$; for any two vertices the squared ratio $c_s^2\mathrm{/}{c}_t^2$ is the product of the edge ratios along a path, and by [F28] $\mathrm{\Gamma }$ contains no cycle, so such a path meets the unique multi-edge at most once and the product equals $1$ when the path avoids the multi-edge and $2^{\pm 1}$ or $3^{\pm 1}$ when it crosses a label-$4$ or label-$6$ edge. Consequently $c_s^2\mathrm{/}{c}_t^2\in \{1,2,2^{-1}\}$ for all $s,t$ when an edge of label $4$ exists, $c_s^2\mathrm{/}{c}_t^2\in \{1,3,3^{-1}\}$ when an edge of label $6$ exists, and $c_s=c_t$ for all $s,t\in S$ when no edge of label $\ge 4$ exists.

This completes all four clauses: (1) is steps 1.1 and 1.2; (2) is step 2.1 together with the ratio alternatives of steps 3.1 and their global form 4.1; (3) is steps 1.5 and 2.2; and (4) is steps 1.6, 2.3, 2.4, 3.2 and 3.3.