4 · Example · Part I of III

\(X=\mathrm{SL}_2/\mathrm{SO}_2\), Part I: the space, in matrices

This part works with matrices only: what \(X\) looks like, how to tell which orbit a loop lies in, why orbits of \(L^+G\) can be read as lines, and the Hecke bound as a valuation count. Part II builds the Bruhat–Tits tree, and Part III explains the same results with it.

1 Why people care about this example

It is the smallest symmetric variety. It has rank one, \(S=T\), and \(\theta\lambda=-\lambda\). It is the case \(n=2\) of \(\mathrm{SL}_n/\mathrm{SO}_n\), a key example of Chen–Yi, and a candidate for the example the draft leaves empty.
It is \(G/T\) for a torus. Over \(\mathbb C\), \(\mathrm{SO}_2\cong\mathbb G_m\) is a maximal torus. Motivation Periods over tori are classical. For the split torus of \(\mathrm{PGL}_2\), the period of a cusp form \(f\) is Hecke's integral \(\int_0^\infty f(iy)\,y^{s-1}\,dy\), which gives the \(L\)-function \(L(s,f)\). Non-split tori lead to Waldspurger's formula. Relative Langlands duality is organised around such period–\(L\)-function pairs.
It is the first place where the relative case differs from the group case. The Hecke operator of size \(m\) moves the orbit index by \(2m\), not \(m\). This doubling is the source of the parity gap in the draft's Sat = Full argument. See the contrast page.

2 Three ways to picture \(X\)

Definition \(X=\{A=\begin{pmatrix}x&y\\y&z\end{pmatrix}:\ xz-y^2=1\}\), with \(g\cdot A=gAg^t\). To each \(A\) attach the quadratic form \(q_A(v)=v^tA^{-1}v\) on the plane.

(a) Real points. \(xz-y^2=1\) is a hyperboloid of two sheets. The sheet \(x>0\), the positive definite forms, is the hyperbolic plane \(\mathrm{SL}_2(\mathbb R)/\mathrm{SO}_2(\mathbb R)\), shown here as a disc. \(\mathrm{SO}_2(\mathbb R)\)-orbits are circles about \(I\): polar coordinates, i.e. the Cartan decomposition \(G=KAK\).

(b) Pairs of points. Over \(\mathbb C\) the form \(q_A\) has two isotropic lines \(\xi_1\neq\xi_2\in\mathbb P^1\). This gives \(X\cong\mathrm{SL}_2/T\cong\{(\xi_1,\xi_2):\xi_1\neq\xi_2\}\), and swapping the pair corresponds to \(A\mapsto-A\). Draw the pair as the endpoints of a geodesic.

Proved here Picture (b). \(\mathrm{SO}_2\) preserves \(x^2+y^2=(x+iy)(x-iy)\), so it fixes its two isotropic points \([\pm i:1]\), i.e. \(x/y=\pm i\). It is exactly the stabilizer of this ordered pair in \(\mathrm{SL}_2\): in the eigenbasis it is the diagonal torus. Writing \(A=gg^t\), the pair of \(A\) is \((g[i:1],\,g[-i:1])\), and \(q_{gAg^t}(v)=q_A(g^{-1}v)\).

(c) Over \(F=\mathbb C(\!(t)\!)\), picture (a) has no analogue, but picture (b) survives. The disc is replaced by the Bruhat–Tits tree, and its boundary circle by \(\mathbb P^1(F)\). A point of \(LX\) is a pair of distinct boundary points of the tree, that is, a line in the tree. Why this is legitimate, and why \(K\)-orbits become distances: step 4. Part II builds the tree, and Part III this picture.

3 Which orbit? Read off the pole order

Definition \(LX=\{A\in\mathrm{Sym}_2(F):\det A=1\}\) with \(g\cdot A=gAg^t\). The pole order of \(A\) is \(-\min_{i,j}\operatorname{val}(A_{ij})\), the largest order of a pole among its entries (\(\operatorname{val}\)).

Proved here \(A\in\mathcal O_a\iff\) the pole order of \(A\) is \(a\). Every \(A\in LX\) is \(K\)-equivalent to exactly one \(\operatorname{diag}(t^a,t^{-a})\), \(a\ge0\).

Proof. For \({\kappa}\in K\) the entries of \({\kappa}A{\kappa}^t\) are \(\mathfrak o\)-combinations of those of \(A\), so the pole order cannot grow; applying \({\kappa}^{-1}\) shows it cannot drop. The representatives \(\operatorname{diag}(t^a,t^{-a})\) of the orbits \(\mathcal O_a\) (Chen–Yi, Prop. 17) have pole order \(a\). \(\square\)

Worked examples

Table 1. \(A\in LX\) and pole order and orbit
\(A\in LX\)	pole order	orbit
\(I\), or any \({\kappa}{\kappa}^t\) with \({\kappa}\in K\)	0	\(\mathcal O_0=L^+X\)
\(\operatorname{diag}(t^{-1},t)=w\operatorname{diag}(t,t^{-1})w^t\), \(w=\begin{pmatrix}0&1\\-1&0\end{pmatrix}\)	1	\(\mathcal O_1\)
\(\begin{pmatrix}1&t^{-1}\\ t^{-1}&1+t^{-2}\end{pmatrix}=uu^t\), \(u=\begin{pmatrix}1&0\\ t^{-1}&1\end{pmatrix}\)	2	\(\mathcal O_2\)
\(\begin{pmatrix}t^{2}&1\\ 1&2t^{-2}\end{pmatrix}\) (determinant \(2-1=1\))	2	\(\mathcal O_2\)

How the pole orders in the table are computed

The pole order is read off entry by entry: take \(\operatorname{val}\) of each entry, take the minimum, change its sign.

Table 2. \(A\) and \(\operatorname{val}\) of the entries \(A_{11},\,A_{12},\,A_{22}\) and min
\(A\)	\(\operatorname{val}\) of the entries \(A_{11},\,A_{12},\,A_{22}\)	min	pole order
\(I\)	\(0,\ \infty,\ 0\)	0	0
\(\operatorname{diag}(t^{-1},t)\)	\(-1,\ \infty,\ 1\)	\(-1\)	1
\(\begin{pmatrix}1&t^{-1}\\ t^{-1}&1+t^{-2}\end{pmatrix}\)	\(0,\ -1,\ -2\) (\(1+t^{-2}=t^{-2}(1+t^2)\))	\(-2\)	2
\(\begin{pmatrix}t^{2}&1\\ 1&2t^{-2}\end{pmatrix}\)	\(2,\ 0,\ -2\)	\(-2\)	2

(\(\operatorname{val}0=\infty\).) The criterion above then says both of the last two lie in \(\mathcal O_2\). To see it by hand, move each to the representative \(\operatorname{diag}(t^2,t^{-2})\) or \(\operatorname{diag}(t^{-2},t^2)\), which lie in one orbit because \(w\in K\) swaps them:

\[ \begin{aligned} {\kappa}&=\begin{pmatrix}1&-t^2/2\\0&1\end{pmatrix}: &{\kappa}\begin{pmatrix}t^{2}&1\\ 1&2t^{-2}\end{pmatrix}{\kappa}^t&=\operatorname{diag}\big(\tfrac12t^2,\ 2t^{-2}\big),\\ {\kappa}&=\begin{pmatrix}0&1\\-1&\frac{t}{1+t^2}\end{pmatrix}: &{\kappa}\begin{pmatrix}1&t^{-1}\\ t^{-1}&1+t^{-2}\end{pmatrix}{\kappa}^t&=\operatorname{diag}\big(\varepsilon\,t^{-2},\ \varepsilon^{-1}t^{2}\big),\quad \varepsilon=1+t^2 . \end{aligned} \]

Both \({\kappa}\) have entries in \(\mathfrak o\) and determinant 1. In each result the diagonal entries are a unit times \(t^{\pm2}\); the unit is a square in \(\mathfrak o\) (\(\sqrt2\in\mathbb C\), \(\sqrt{1+t^2}=1+\tfrac12t^2-\cdots\)), so \(\operatorname{diag}(s,s^{-1})\in K\) removes it.

The third row is a whole family: replacing \(t^{-1}\) by \(s\,t^{-1}\) gives a point of \(\mathcal O_2\) for \(s\neq0\) and \(I\in\mathcal O_0\) at \(s=0\). So \(\mathcal O_0\subset\overline{\mathcal O}_2\) (orbit closures).

4 Why orbits of \(L^+G\) on \(LX\) can be read as lines

Uses later pages. The tree (definition; built in Part II), its ends (Part II, step 5), the stabilizer of \(v_0\) (Part II, step 6), and the ends of \(\ell_A\) (Part III, step 2). On a first reading, read the short answer only and return after Part III.

Short answer Strictly, a point of \(LX\) is a line of the tree, and an orbit of \(K=L^+G\) is the set of all lines at one distance from \(v_0\): \[ LX\ \xrightarrow{\ \sim\ }\ \{\text{oriented lines of the tree}\},\qquad \mathcal O_a=\{\ell:\ \operatorname{dist}(v_0,\ell)=a\}. \]

Rule of thumb. Draw a point of \(G/H\) as the part of the tree that its stabilizer \(H\) preserves.
The two homogeneous spaces. In \(\mathrm{Gr}=G/K\), the stabilizer \(K\) fixes one vertex, so a point is a vertex. In \(LX=G/\mathrm{SO}_2(F)\), the stabilizer is a split torus, which fixes no vertex but preserves exactly one line, its apartment, and slides along it. So a point is a line.
The orbits. \(K\) fixes \(v_0\), so the only thing it cannot change about a line is the line's distance to \(v_0\).

Table 3. space and stabilizer of the base point and what it preserves
space	stabilizer of the base point	what it preserves	so a point is
\(\mathrm{Gr}=\mathrm{SL}_2(F)/K\)	\(K=\mathrm{SL}_2(\mathfrak o)\)	the vertex \(v_0\)	a vertex (an even one)
\(LX=\mathrm{SL}_2(F)/\mathrm{SO}_2(F)\)	\(\mathrm{SO}_2(F)\cong F^\times\), a split torus	the line \(\ell_I\) from \(-i\) to \(i\)	an oriented line
over \(\mathbb R\): \(\mathrm{SL}_2(\mathbb R)/\mathrm{SO}_2(\mathbb R)\)	\(\mathrm{SO}_2(\mathbb R)\), compact	the centre of the disc	a point of the disc, picture (a)

The last row is why the real picture (a) has no analogue over \(F\). Over \(\mathbb R\), \(x^2+y^2\) has no isotropic line and \(\mathrm{SO}_2(\mathbb R)\) is a circle. Over \(F\supset\mathbb C\), \(x^2+y^2=(x+iy)(x-iy)\) has two isotropic lines and \(\mathrm{SO}_2(F)\) is a copy of \(F^\times\) (Casselman's notes open with this analogy between the hyperbolic disc and the tree [Casselman, Geometry of the tree, p. 1]). The five steps below make the short answer precise.

Proved here (i) \(LX\) is one orbit: \(g\mapsto gg^t\) identifies \(\mathrm{SL}_2(F)/\mathrm{SO}_2(F)\) with \(LX\).

Proof. Let \(A\in LX\). The binary form \(q_A\) has determinant 1, and \(-1=i^2\) is a square in \(F\), so \(q_A\) is isotropic. A non-degenerate isotropic binary form is a hyperbolic plane, and so is \(x^2+y^2\). So \(A=gg^t\) for some \(g\in\mathrm{GL}_2(F)\). Then \(\det g=\pm1\), and if \(\det g=-1\) we replace \(g\) by \(g\operatorname{diag}(1,-1)\). The stabilizer of \(I\) is \(\{g:gg^t=I,\ \det g=1\}=\mathrm{SO}_2(F)\). \(\square\)

Proved here (ii) \(\mathrm{SO}_2(F)\) is the split torus with eigenlines \(\ell_\pm=F\cdot(\pm i,1)\). That is, \(\mathrm{SO}_2(F)=\{g\in\mathrm{SL}_2(F):\ g\ell_+=\ell_+,\ g\ell_-=\ell_-\}\cong F^\times\).

Proof. \(\begin{pmatrix}a&-b\\b&a\end{pmatrix}\begin{pmatrix}\pm i\\1\end{pmatrix}=(a\pm ib)\begin{pmatrix}\pm i\\1\end{pmatrix}\). Conversely, with \(P=\begin{pmatrix}i&-i\\1&1\end{pmatrix}\), every \(g\in\mathrm{SL}_2(F)\) with eigenlines \(\ell_\pm\) is \(P\operatorname{diag}(\lambda,\lambda^{-1})P^{-1}=\begin{pmatrix}a&-b\\b&a\end{pmatrix}\), where \(a=\tfrac12(\lambda+\lambda^{-1})\), \(b=\tfrac1{2i}(\lambda-\lambda^{-1})\) and \(a^2+b^2=1\). \(\square\)

Proved (cited) (iii) A split torus has a line, its apartment. Let \(x_0,x_\infty\) span two distinct lines of \(F^2\), which are two ends of the tree. The geodesic between these ends has vertices \([\mathfrak o\,t^mx_0+\mathfrak o\,x_\infty]\), \(m\in\mathbb Z\). It is carried into itself by the matrices with eigenlines \(Fx_0\) and \(Fx_\infty\), and the torus acts on it by translations [Casselman, Geometry of the tree, Cor. 7.2; ends are lines of \(F^2\) by Prop. 5.4]. The same lattice description of an apartment is Joswig–Sturmfels–Yu's, over \(\mathbb C(\!(z)\!)\) exactly [JSY, Affine buildings and tropical convexity, §2].

Instance. For \(A=I\), take \(x_0=(i,1)\) and \(x_\infty=(-i,1)\). The vertex \(m=0\) is \(\mathfrak o(i,1)+\mathfrak o(-i,1)=\mathfrak o^2\), since \(\det\begin{pmatrix}i&-i\\1&1\end{pmatrix}=2i\) is a unit. So \(\ell_I\) passes through \(v_0\), as it must, since \(I\in\mathcal O_0\). The element \(\lambda=t\) of the torus moves \(m\) to \(m+2\): it translates \(\ell_I\) by 2.

Proved here (iv) Points are lines. \(\Phi(gg^t)=g\,\ell_I\), oriented from \(g(-i)\) to \(g(i)\), is a bijection from \(LX\) to the oriented lines, and \(\Phi(h\cdot A)=h\,\Phi(A)\).

Proof. Well defined: if \(gg^t=g'g'^t\), then \(g^{-1}g'\in\mathrm{SO}_2(F)\) fixes both ends of \(\ell_I\) by (ii). Injective: in a tree a line is determined by its two ends. So if \(g\ell_I=g'\ell_I\) with the same orientation, \(g^{-1}g'\) fixes \(\ell_+\) and \(\ell_-\), lies in \(\mathrm{SO}_2(F)\) by (ii), and \(gg^t=g'g'^t\). Surjective: given distinct ends \(F v_1, F v_2\), rescale \(v_2\) so that \(\det(v_1,v_2)=2i\). Then \(g=(v_1\ v_2)P^{-1}\in\mathrm{SL}_2(F)\) sends \((i,1)\mapsto v_1\) and \((-i,1)\mapsto v_2\). Equivariant: \(h\cdot gg^t=(hg)(hg)^t\). \(\square\) Part III proves the same bijection by formulas: the ends of \(\Phi(A)\) are \(\xi_\pm=(y\pm i)/z\).

Proved here (v) \(K\)-orbits are distances. \(\Phi(\mathcal O_a)=\{\ell:\operatorname{dist}(v_0,\ell)=a\}\), and \(\Phi(\overline{\mathcal O}_a)\) is the set of lines that meet the ball of radius \(a\) about \(v_0\).

Proof. \(K\) is the stabilizer of \(v_0\) (Part II, step 6) and acts by isometries, so \(\operatorname{dist}(v_0,\Phi(A))\) is constant on each \(K\)-orbit. By step 3 the orbits are the \(\mathcal O_a\), with representatives \(\operatorname{diag}(t^a,t^{-a})\). The ends of \(\Phi(\operatorname{diag}(t^a,t^{-a}))\) are \(\pm i\,t^a\) (Part III, step 2). These two power series share their first \(a\) Taylor coefficients, so the line is at distance \(a\) from \(v_0\) (Part II, step 5). Since \(\Phi\) is a bijection and the \(\mathcal O_a\) cover \(LX\), the lines at distance \(a\) are exactly \(\Phi(\mathcal O_a)\); the closure statement follows from \(\overline{\mathcal O}_a=\bigsqcup_{b\le a}\mathcal O_b\). \(\square\)

Where it fails The argument needs \(i=\sqrt{-1}\in k\). Over \(k=\mathbb F_q\) with \(q\equiv3\bmod4\), \(x^2+y^2\) has no isotropic line in \(F^2\), so \(\mathrm{SO}_2(F)\) has no eigenline, fixes no end, and points of \(X(F)\) are not lines. This is the non-split torus of row 3 of the table again, now over \(F\).

5 The Hecke bound, by counting poles

Proved here If \(g\in D_{\le m}\) (entries with poles of order \(\le m\)) and \(A\in\overline{\mathcal O}_a\), then each entry \((gAg^t)_{ij}=\sum_{k,l}g_{ik}A_{kl}g_{jl}\) has a pole of order at most \(m+a+m\). Hence \[ D_{\le m}\cdot\overline{\mathcal O}_a\subset\overline{\mathcal O}_{a+2m}. \] The bound is attained: \(g=t^{(m,-m)}\) and \(A=I\) give \(gg^t=\operatorname{diag}(t^{2m},t^{-2m})\in\mathcal O_{2m}\).

Instance: \(m=1\), \(A=I\)

\(g=\operatorname{diag}(t,t^{-1})\) and \(g=u\) (row 3 above) both land in \(\mathcal O_2\), and \(g\in K\) lands in \(\mathcal O_0\). Does \(D_{\le1}\cdot I\) also meet \(\mathcal O_1\)? The count \(m+a+m\) cannot say. Question 1 asks, and Part III answers it with a picture.

Heuristic The count shows that the index grows by \(2m\), but not why it is \(2m\) rather than \(m\), nor which orbits below the top are reached. Part III shows both: \(g\) acts on both ends of a line.