Hg toolkit finitely many cayley cone types

We first derive the radial bound needed below. For specified geodesics $[o,x],[o,y]$, suppose $0\le t\le (x\mathrm{\mid }y{)}_o$, and let $x_t,y_t$ be their radius-$t$ points. They exist since products do not exceed either radial length. We have $(x_t\mathrm{\mid }x{)}_o=t$ and $(y\mathrm{\mid }{y}_t{)}_o=t$. Apply F2 through the chain $x_t,x,y,y_t$: two product inequalities give $(x_t\mathrm{\mid }{y}_t{)}_o\ge t-2\kappa$. Hence $d(x_t,y_t)=2t-2(x_t\mathrm{\mid }{y}_t{)}_o\le 4\kappa =12\delta$. This calculation includes $t=0$ and $\delta =0$.

Fix $g,g^{\mathrm{\prime }}$ with $P_r(g)=P_r(g^{\mathrm{\prime }})$. The identity lies in both cones. If $\mathrm{\mid }h\mathrm{\mid }=1$ and $h\in \mathcal{C}(g)$, then $h\mathrm{\notin }{P}_r(g)$ and hence $h\mathrm{\notin }{P}_r(g^{\mathrm{\prime }})$. Thus $\mathrm{\mid }{g}^{\mathrm{\prime }}h\mathrm{\mid }>\mathrm{\mid }{g}^{\mathrm{\prime }}\mathrm{\mid }$, while $\mathrm{\mid }{g}^{\mathrm{\prime }}h\mathrm{\mid }\le \mathrm{\mid }{g}^{\mathrm{\prime }}\mathrm{\mid }+1$. Integral word lengths force equality, proving $h\in \mathcal{C}(g^{\mathrm{\prime }})$.

Induct on $l=\mathrm{\mid }h\mathrm{\mid }\ge 2$, assuming cone membership transfers for shorter elements. If $h\in \mathcal{C}(g)$, take a shortest spelling $h=h^{\mathrm{\prime }}s$ with $\mathrm{\mid }{h}^{\mathrm{\prime }}\mathrm{\mid }=l-1$ and $\mathrm{\mid }s\mathrm{\mid }=1$. The inequalities $\mathrm{\mid }g\mathrm{\mid }+l=\mathrm{\mid }gh\mathrm{\mid }\le \mathrm{\mid }g{h}^{\mathrm{\prime }}\mathrm{\mid }+1\le \mathrm{\mid }g\mathrm{\mid }+l$ force $h^{\mathrm{\prime }}\in \mathcal{C}(g)$; hence $\mathrm{\mid }{g}^{\mathrm{\prime }}{h}^{\mathrm{\prime }}\mathrm{\mid }=\mathrm{\mid }{g}^{\mathrm{\prime }}\mathrm{\mid }+l-1$ by induction. Suppose for a contradiction that $h\mathrm{\notin }\mathcal{C}(g^{\mathrm{\prime }})$. Writing $A=\mathrm{\mid }{g}^{\mathrm{\prime }}\mathrm{\mid }$, the endpoint $g^{\mathrm{\prime }}h$ has length between $A+l-2$ and $A+l-1$, by its distance $1$ from $g^{\mathrm{\prime }}{h}^{\mathrm{\prime }}$ and the failed cone equality. In particular its length is at least $A$.

Choose the geodesic from $e$ through $g^{\mathrm{\prime }}$ to $g^{\mathrm{\prime }}{h}^{\mathrm{\prime }}$ provided by the equality in step 2.1, and any geodesic from $e$ to $g^{\mathrm{\prime }}h$. Their endpoints have distance $1$ and product at least $((A+l-1)+(A+l-2)-1)\mathrm{/}2=A+l-2\ge A$. Let $k_1$ be the vertex at radius $A$ on the second geodesic, and put $k_2=k_1^{-1}{g}^{\mathrm{\prime }}h$. The vertex exists because its length is at least $A$. Step 1.1 gives $\mathrm{\mid }{g}^{\mathrm{\prime }-1}{k}_1\mathrm{\mid }=d(g^{\mathrm{\prime }},k_1)\le 12\delta \le r$. Also $\mathrm{\mid }{k}_1\mathrm{\mid }=A$ and $\mathrm{\mid }{k}_2\mathrm{\mid }=\mathrm{\mid }{g}^{\mathrm{\prime }}h\mathrm{\mid }-A\le l-1$. Set $v=g^{\mathrm{\prime }-1}{k}_1$. Then $v\in P_r(g^{\mathrm{\prime }})=P_r(g)$, so $\mathrm{\mid }gv\mathrm{\mid }\le \mathrm{\mid }g\mathrm{\mid }$.

Since $h=v{k}_2$, we obtain $\mathrm{\mid }g\mathrm{\mid }+l=\mathrm{\mid }gh\mathrm{\mid }\le \mathrm{\mid }gv\mathrm{\mid }+\mathrm{\mid }{k}_2\mathrm{\mid }\le \mathrm{\mid }g\mathrm{\mid }+l-1$, a contradiction. This proves $\mathcal{C}(g)\subseteq \mathcal{C}(g^{\mathrm{\prime }})$ by induction. Interchanging $g,g^{\mathrm{\prime }}$ proves equality. The finite-radius set in the statement is finite by F1 (it is contained in the open radius-$r+1$ ball), so has finitely many subsets, at most $2$ to its cardinality. Each possible $P_r(g)$ determines just one cone, establishing finiteness and the stated count. Only finitely many geodesics are chosen for each specified inductive comparison; no AC is used.