The lattice tree and Iwahori orbits for SL2(Q3)

First derive the residue and valuation consequences needed below from [F2]. If $0\mathrm{\ne }x\in \mathcal{O}$, its leading exponent $r$ is nonnegative because $\mathrm{\mid }x{\mathrm{\mid }}_3\le 1$; the expansion gives $x=a_0+\pi y$ with $a_0\in \{0,1,2\}$ and $y\in \mathcal{O}$, since the shifted tail is a limit of partial sums in the closed unit ball. Distinct such digits represent distinct residue classes, since a nonzero difference in $\{\pm 1,\pm 2\}$ has absolute value $1$, whereas every element of $\pi \mathcal{O}$ has absolute value at most $1\mathrm{/}3$. Thus $\mathcal{O}\mathrm{/}\pi \mathcal{O}$ has exactly the three classes $0,1,2$ with the operations of ${\mathbb{F}}_3$. For a nonzero $x\in \mathcal{O}$ with leading exponent $r$, the shifted expansion gives $x={\pi }^{r}u$ with $u=a_r+\pi z$, $a_r\in \{1,2\}$ and $z\in \mathcal{O}$. The leading digit gives $\mathrm{\mid }u{\mathrm{\mid }}_3=\mathrm{\mid }{u}^{-1}{\mathrm{\mid }}_3=1$, so [F1] gives $u,u^{-1}\in \mathcal{O}$ and $u$ is a unit. Every nonzero ideal of $\mathcal{O}$ therefore contains ${\pi }^r$ for the minimum valuation $r$ of its elements, and every element of that ideal is divisible by ${\pi }^r$ in $\mathcal{O}$. Thus every ideal is principal and $\mathcal{O}$ is a PID. After scaling any lattice $M$ into ${\mathcal{O}}^2$, [F3] gives an $\mathcal{O}$-basis $f_1,f_2$ of ${\mathcal{O}}^2$ with $M=\mathcal{O}{\pi }^a{f}_1+\mathcal{O}{\pi }^b{f}_2$, where $0\le a\le b$ after absorbing unit factors into the basis. Scaling to $a=0\le b$ gives a primitive representative and an exponent $n=b-a$. This exponent is well defined on the homothety class: a primitive representative has minimum coordinate valuation zero, so any scalar multiple that is also primitive differs by an $\mathcal{O}$-unit. For a basis matrix of the primitive representative, the determinant valuation is $n$; changing either $\mathcal{O}$-basis multiplies the determinant by a unit. Neighbors of $[L]$ correspond to the four lines in $L\mathrm{/}\pi L\cong {\mathbb{F}}_3^2$, because the inverse image of a line is an intermediate lattice. Distinct lines give distinct lattices, and they cannot become homothetic: all four inverse images have the same index $3$ in $L$, so a scalar relating two of them has valuation zero and is a unit, which preserves each lattice.

Relative to the base vertex $[{\mathcal{O}}^2]$, a vertex with normalized elementary divisors $(0,n)$ has a chain of length $n$ through $\mathcal{O}{f}_1+{\pi }^j\mathcal{O}{f}_2$. If $n>0$, exactly one of its four neighbors has distance parameter $n-1$: in the basis $f_1,{\pi }^n{f}_2$ of its representative, the line spanned by ${\pi }^n{f}_2$ has inverse image $\pi \mathcal{O}{f}_1+{\pi }^n\mathcal{O}{f}_2$, whose homothety class normalizes to parameter $n-1$. Every other line has inverse image with a primitive vector $f_1+t{\pi }^n{f}_2$ and second elementary divisor $n+1$, hence parameter $n+1$. At $n=0$ all neighbors have parameter $1$. Therefore the parameter changes by one at each step, and the displayed chain is a shortest path. A cycle would have a vertex of maximal parameter and two distinct adjacent vertices of parameter one less, contradicting uniqueness of the decreasing neighbor. Thus the graph is connected and has no cycle.

Replacing a lattice basis multiplies its determinant by an $\mathcal{O}$-unit; homothety adds twice an integer to its determinant valuation. Hence type is well defined, and determinant-one matrices preserve it. For an edge with the endpoint of type $0$ first, choose a basis $f_1,f_2$ of its first lattice with second lattice $\mathcal{O}{f}_1+\pi \mathcal{O}{f}_2$. Its determinant has even valuation. Rescale both vectors by a power of $\pi$ to make that determinant a unit and then multiply $f_1$ by its inverse; neither operation changes the edge, and the resulting change-of-basis matrix has determinant one. This proves transitivity. A determinant-one matrix fixes $[L_0]$ exactly when it preserves $L_0$ (homothety would change determinant valuation unless its scalar is a unit). Within ${\mathrm{SL}}_2(\mathcal{O})$, preservation of $L_1$ is exactly $c\in \pi \mathcal{O}$, proving the stated stabilizer.

The unique shortest edge gallery from $E$ to any other edge changes endpoint types alternately: changing twice at the same endpoint would shorten it. For a gallery proceeding along $L_0,L_1,\dots ,L_n$, the matrices $\left(\begin{array}{cc}1& 0\\ {\pi }^{n}b& 1\end{array}\right)$ for $b\in \mathcal{O}$ fix every earlier vertex and act transitively on the three possible next edges. In the basis $e_1,{\pi }^n{e}_2$, the backward edge corresponds to the line spanned by the second basis vector; the other lines are spanned by $e_1+t{\pi }^n{e}_2$ for $t\in {\mathbb{F}}_3$, and the displayed matrices send $t$ to $t+b$. For the other initial direction the apartment chain is $L_1,L_0,L_{-1},\dots ,L_{-n}$; use instead $\left(\begin{array}{cc}1& {\pi }^{n}b\\ 0& 1\end{array}\right)$ at $L_{-n}$, which fixes that earlier chain and is transitive on the three forward lines there. Inductively move each gallery onto the corresponding apartment segment while fixing its earlier segment. Hence $I$ is transitive on the edges with each alternating type word. There are three choices at each extension, so the orbit size is $3^n$. Different words have different type sequences or distances and cannot be in the same orbit. By step 2.2 edges identify with $G\mathrm{/}I$, so these orbits are exactly the double cosets.