7 · Example · contrast
The group \(\mathrm{SL}_2\): same orbits, no doubling
This is the classical case of geometric Satake. Its orbit poset is the same as in the running example, but its Hecke map is different. That difference is why odd orbits are never top orbits in the running example (Part III).
1 Setup
Definition
\(G'=\mathrm{SL}_2\times\mathrm{SL}_2\) acts on \(X=\mathrm{SL}_2\) by \((g,h)\cdot x=g\,x\,h^{-1}\). This is the symmetric variety \(G'/\Delta\mathrm{SL}_2\) for \(\theta(g,h)=(h,g)\), with \(\tau(g,h)\) identified with \(gh^{-1}\). The arc group is \(K\times K\), and for \(\lambda=((m_1,-m_1),(m_2,-m_2))\) one has \(D_{\le\lambda}=D_{\le m_1}\times D_{\le m_2}\).
Proved (cited)
The \(K\times K\)-orbits on \(LX=\mathrm{SL}_2(F)\) are the Cartan double cosets \(\mathcal O_a:=Kt^{(a,-a)}K\), and \(\overline{\mathcal O}_a=\mathcal O_0\sqcup\dots\sqcup\mathcal O_a\) [Zhu, Prop. 2.1.5]. The index \(a\) is the pole order of \(x\), exactly as for \(\mathrm{SL}_2/\mathrm{SO}_2\).
2 The picture: points are vertices
Proved here
Send \(x\) to the vertex \(xv_0\). Then \(({\kappa}_1,{\kappa}_2)\cdot x\) goes to \({\kappa}_1xv_0\), at the same distance from \(v_0\), and by Serre's criterion (Part II, step 6)
\[
x\in\mathcal O_a\iff \operatorname{dist}(v_0,xv_0)=2a .
\]
3 Hecke translation
Proved here
For \(g\in D_{\le m_1}\), \(h\in D_{\le m_2}\) and \(x\in\overline{\mathcal O}_a\), apply the triangle inequality twice:
\[
\operatorname{dist}(v_0,\,gxh^{-1}v_0)\le \underbrace{\operatorname{dist}(v_0,gv_0)}_{\le 2m_1}+\underbrace{\operatorname{dist}(v_0,xv_0)}_{\le 2a}+\underbrace{\operatorname{dist}(xv_0,xh^{-1}v_0)}_{=\operatorname{dist}(v_0,h^{-1}v_0)\,\le 2m_2},
\]
so \((D_{\le m_1}\times D_{\le m_2})\cdot\overline{\mathcal O}_a\subset\overline{\mathcal O}_{a+m_1+m_2}\).
Every orbit is a top. Take \(\lambda=((a,-a),(0,0))\). Then \(t^\lambda\cdot x_0=t^{(a,-a)}\in\mathcal O_a\) for every \(a\), odd or even.
4 Side by side
| a point of \(LX\) is | Hecke of size \(\lambda\) moves the index by | tops of \(D_{\le\lambda}\cdot x_0\) | |
|---|---|---|---|
| \(\mathrm{SL}_2/\mathrm{SO}_2\) | a line \(\ell_A\) | \(2m\) (one \(g\), on both sides) | even \(a\) only |
| group \(\mathrm{SL}_2\) | a vertex \(xv_0\) (index = half the distance) | \(m_1+m_2\) (two independent elements) | every \(a\) |
Both cases have the orbit poset \(0<1<2<\cdots\). The difference is the Hecke map \(\lambda\mapsto\overline\lambda\). In the group case \(g\) and \(h\) are independent, so the two sides of \(x\) can move by different amounts. In the symmetric case \(\tau(g\cdot x)=g\,\tau(x)\,\theta(g)^{-1}\) uses one \(g\) on both sides.