6 · Example · Part III of III

\(X=\mathrm{SL}_2/\mathrm{SO}_2\), Part III: points of \(LX\) are lines in the tree

Part I sorts \(A\in LX\) by the pole order of its entries. Part II builds the tree, whose ends are \(\mathbb P^1(F)\). This part draws each \(A\) as a line in the tree, then reads the Hecke bound off the picture.

1 Why a matrix has two zeros, and what they are for

Goal. Draw a point \(A\in LX\) in the tree, so that its orbit becomes a distance and a Hecke operator becomes a triangle inequality (step 5). The tree has no vertex for \(A\), but its are \(\mathbb P^1(F)\), and \(A\) determines exactly two of them.

Why exactly two. Since \(i\in F\), \(\mathrm{SO}_2\) is the stabilizer of the ordered pair \((i,-i)\) under Möbius transformations, so \[ X=\mathrm{SL}_2/\mathrm{SO}_2\ \xrightarrow{\ \sim\ }\ \{(\xi_+,\xi_-)\in\mathbb P^1(F)^2:\ \xi_+\neq\xi_-\},\qquad A=gg^t\ \longmapsto\ \big(g\cdot i,\ g\cdot(-i)\big). \] The pair is read off \(A\) without choosing \(g\): it is the two (zeros) of \(q_A\) (step 2). Nothing is lost: the pair determines \(A\) (step 4, T0).

\(A=\begin{pmatrix}x&y\\y&z\end{pmatrix}\in LX\)symmetric, \(\det A=1\)
↓\(q_A(v)=v^tA^{-1}v\)
binary quadratic form \(q_A\)
↓its two zeros (step 2)
\(\xi_\pm=\dfrac{y\pm i}{z}\in\mathbb P^1(F)\)
↓\(\mathbb P^1(F)=\) ends (step 3)
two ends of the treetwo rays from \(v_0\)
↓join them (step 4)
the \(\ell_A\)
Figure 1. A point of \(LX\) becomes a line in the tree through the two zeros of its quadratic form.

Payoff: the three ways of measuring \(A\) agree,

\[ \underbrace{\text{pole order of }A}_{\text{Part I}}\ =\ \underbrace{\operatorname{val}(\xi_+-\xi_-)}_{\text{shared Taylor coefficients}}\ =\ \underbrace{\operatorname{dist}(v_0,\ell_A)}_{\text{Part II}}\qquad(\xi_\pm\in\mathfrak o). \]

Same move as picture (b) of Part I: over \(\mathbb C\) a point of \(X\) is a pair of boundary points; over \(F\) the boundary is \(\mathbb P^1(F)\) around the tree.

2 A matrix gives two points of \(\mathbb P^1(F)\)
Definition For \(A=\begin{pmatrix}x&y\\y&z\end{pmatrix}\in LX\), \(A^{-1}=\begin{pmatrix}z&-y\\-y&x\end{pmatrix}\), so \[ q_A(v_1,v_2)=z\,v_1^2-2y\,v_1v_2+x\,v_2^2 . \] A point \([v_1:v_2]\in\mathbb P^1(F)\) is for \(q_A\) when \(q_A(v_1,v_2)=0\). We write a point with \(v_2\neq0\) by its coordinate \(\xi=v_1/v_2\in F\).
Proved here \(q_A\) has exactly two isotropic points, the ordered pair \[ \xi_+=[\,y+i:z\,],\qquad \xi_-=[\,y-i:z\,],\qquad \xi_\pm=\frac{y\pm i}{z} . \] (If \(z=0\), then \(y=\pm i\) and the two points are \(\infty\) and \(x/2y\).) They are distinct and \(F\)-rational, because \(i=\sqrt{-1}\in\mathbb C\subset F\). If \(A=gg^t\), then \(\xi_\pm=g\cdot(\pm i)\), with \(g\) acting on \(\mathbb P^1(F)\) by Möbius transformations.
? Proof
Put \(v=(y\pm i,z)\): \(q_A(v)=z\big[(y\pm i)^2-2y(y\pm i)+xz\big]=z\big[-(1+y^2)+xz\big]=0\), using \(xz-y^2=1\). They are distinct because \(\det\begin{pmatrix}y+i&y-i\\z&z\end{pmatrix}=2iz\neq0\). If \(z=0\), then \(q_A=v_2(x\,v_2-2y\,v_1)\). For \(A=I\) the points are \(\pm i\), and \(q_{gAg^t}(v)=q_A(g^{-1}v)\) moves them by \(g\). \(\square\)
Table 1 · Instances of the two zeros
Table 1. \(A\) and \(\xi_+,\ \xi_-\)
\(A\)\(\xi_+,\ \xi_-\)
\(I\)\(i,\ -i\)
\(\operatorname{diag}(t^a,t^{-a})\)\(i\,t^a,\ -i\,t^a\)
\(\begin{pmatrix}1&t^{-1}\\t^{-1}&1+t^{-2}\end{pmatrix}\)\(\dfrac{t}{1-it},\ \dfrac{t}{1+it}\)
\(-A\)the pair of \(A\), swapped

3 Two ends, one line: the shared trunk is the distance

By Part II, step 5, a point \(\xi\in\mathfrak o\) is an end of the tree: the ray from \(v_0\) through the vertices \(\xi\bmod t^n\). A point with a pole is read in the chart at \(\infty\), \(w=1/\xi\in t\mathfrak o\). The line \(\ell\) joining two ends is the union of the two rays minus their shared part.

Proved here For ends \(\xi_+\neq\xi_-\) in \(\mathfrak o\), the rays share exactly the vertices \(\xi_\pm\bmod t^n\) with \(n\le\operatorname{val}(\xi_+-\xi_-)\), so \[ \operatorname{dist}(v_0,\ell)=\operatorname{val}(\xi_+-\xi_-), \] the number of leading Taylor coefficients the two ends share. Chart-free form. If \(p_\pm\in\mathfrak o^2\) are primitive vectors (not both entries divisible by \(t\)) spanning the two ends, then \(\operatorname{dist}(v_0,\ell)=\operatorname{val}\det(p_+,p_-)\).
? Proof
The first statement is Part II, step 5. For the second, \(K\) acts transitively on primitive vectors and preserves both sides, so take \(p_+=(1,0)\), the end \(\infty\), and \(p_-=(a,b)\). If \(\operatorname{val}b=0\), the two lines differ mod \(t\), the rays leave \(v_0\) along different edges, and the distance is 0. If \(\operatorname{val}b>0\), then \(a\) is a unit, \(p_-\) is the point \(w=b/a\) of the chart at \(\infty\), and the rays share \(\operatorname{val}(w-0)=\operatorname{val}b\) steps. \(\square\)
Figure 2. The rays of \(\xi_+,\xi_-\) from \(v_0\): the shaded trunk (labels \(\xi\bmod t^n\)) is where they agree, and its length is \(\operatorname{dist}(v_0,\ell_A)\). Past the ring, the closest point of \(\ell_A\) to \(v_0\), each vertex shows the coefficient it adds; the table shades the shared coefficients.

4 Points of \(LX\) are lines, and orbits are distances
Definition For \(A\in LX\), let \(\ell_A\) be the line in the tree from \(\xi_-\) to \(\xi_+\) (step 2), oriented towards \(\xi_+\).
Proved here (T0) \(A\mapsto\ell_A\) is a bijection from \(LX\) to the oriented lines of the tree.
(T1) \(\ell_{g\cdot A}=g\,\ell_A\) for \(g\in\mathrm{SL}_2(F)\).
(T2) \(\text{pole order of }A=\operatorname{dist}(v_0,\ell_A)\). So \(\mathcal O_a\) is the set of lines at distance \(a\) from \(v_0\), and \(\overline{\mathcal O}_a\) is the set of lines that meet the ball of radius \(a\).
? Proof
(T1) follows from \(q_{gAg^t}(v)=q_A(g^{-1}v)\). (T0): if \(A,A'\) have the same ends, then \(q_{A'}=c\,q_A\), so \(A'^{-1}=cA^{-1}\) and \(c=\pm1\) by determinants; \(c=-1\) swaps the ends (table above), so \(A'=A\). Conversely, \(\mathrm{SL}_2(F)\) is transitive on ordered pairs of distinct points of \(\mathbb P^1(F)\), so some \(g\) sends \((i,-i)\) to a given pair, and \(A=gg^t\) has that pair.
(T2), directly. Write \(A=gg^t\), \(p=g\,(i,1)\), \(q=g\,(-i,1)\). Then \(A=\tfrac12(pq^t+qp^t)\) and \(\det(p,q)=2i\). Factor \(p=t^\alpha p'\), \(q=t^\beta q'\) with \(p',q'\) primitive. Step 3 gives \(\operatorname{dist}(v_0,\ell_A)=\operatorname{val}\det(p',q')=-\alpha-\beta\). And \(A=t^{\alpha+\beta}\cdot\tfrac12(p'q'^t+q'p'^t)\), and the last matrix has an entry that is a unit: otherwise, mod \(t\), \(\bar p'\bar q'^t\) would be a rank-one matrix equal to minus its transpose, which is impossible. So the pole order of \(A\) is \(-\alpha-\beta\) too. \(\square\)
? Second proof of (T2), via the orbit representatives

Both sides are \(K\)-invariant: \(K\) fixes \(v_0\), and \(\operatorname{dist}(v_0,{\kappa}\ell)=\operatorname{dist}({\kappa}^{-1}v_0,\ell)\). By Part I, step 3 it suffices to check \(A=\operatorname{diag}(t^a,t^{-a})\), with pole order \(a\). Its ends are \(\pm i\,t^a\) (step 2), and \(\operatorname{val}(2i\,t^a)=a\) (step 3).

Line at distance:
Figure 3. The line \(\ell\) (blue) lies in \(\mathcal O_a\), where \(a\) is the radius of the shaded ball it just touches at the ring; click two outer vertices to choose its ends. Filled / hollow vertices: even / odd distance from \(v_0\).

Observe. Lines exist at every distance \(a=0,1,2,\dots\), including odd ones, although only even vertices are of the form \(gv_0\). The orbit \(\mathcal O_1\) is the set of lines that just miss \(v_0\) by passing through one of its neighbours.

5 Hecke translation is the triangle inequality

To compute \(g\cdot\ell\), you can either move the line, or move the observer the other way: \(\operatorname{dist}(v_0,g\ell)=\operatorname{dist}(g^{-1}v_0,\ell)\).

Proved here (T3) If \(g\in D_{\le m}\) (loops with poles of order \(\le m\)) and \(A\in\overline{\mathcal O}_a\), then \[ \operatorname{dist}(v_0,\ell_{g\cdot A})=\operatorname{dist}(g^{-1}v_0,\ell_A)\ \le\ \underbrace{\operatorname{dist}(g^{-1}v_0,v_0)}_{\le 2m}+\underbrace{\operatorname{dist}(v_0,\ell_A)}_{\le a}, \] \[ \text{so}\qquad D_{\le m}\cdot\overline{\mathcal O}_a\subset\overline{\mathcal O}_{a+2m}. \] Equality occurs when the observer walks straight away from the line. For example, \(g=t^{(m,-m)}\) and \(A=I\) give \(gg^t=\operatorname{diag}(t^{2m},t^{-2m})\in\mathcal O_{2m}\).
Figure 4. Pick \(m\), then pick \(g\in D_{\le m}\) by clicking its observer \(g^{-1}v_0\), a filled vertex inside the dashed circle of radius \(2m\). Each such vertex is labelled by \(a'=\operatorname{dist}(g^{-1}v_0,\ell_A)\), the orbit of \(g\cdot A\); the table collects these values, and none exceeds \(a+2m\).

The same bound by counting poles is Part I, step 5. The tree adds the reason for \(2m\): \(g\) acts on both ends of the line, so the observer moves by \(2m\).

6 What to notice
  1. The factor 2. A Hecke operator of size \(m\) moves the observer by \(2m\) edges, because \(\mathrm{SL}_2\) moves \(v_0\) only by even distances. So the top index grows by \(2m\).
  2. Everything below the top is reached. Start with \(a=0\) and put the observer at distance 2. Its distance to the line can be 0 (observer on \(\ell\)), 1 (one step off \(\ell\)) or 2. So \(D_{\le1}\cdot x_0\) meets \(\mathcal O_0,\mathcal O_1,\mathcal O_2\).
  3. Odd orbits are never tops. From a line through \(v_0\), the farthest the observer can get is \(2m\), which is even. So \(\mathcal O_1,\mathcal O_3,\dots\) are never the top orbit of \(D_{\le\lambda}\cdot x_0\). An induction that produces sheaves only on top orbits, like the draft's Sat = Full argument, never reaches \(\mathcal O_1\) directly. The contrast page shows the group case has no such parity.