Functions, \(\phiz\) and \(\HK\)
The left-hand side of the Theorem: a space of functions on the \(K\)-orbits of the Geometry tab, a distinguished element \(\phiz\), and the algebra \(\HK\) that acts on it. The question the Theorem answers, whether \(\phiz\) generates, can be tried by hand in the lab below before any Satake transform appears.
The space \(\CK\) and the basic functionWhich functions does the Theorem describe?
Definition
\[\Sw(X):=C_c^\infty(X(F)),\qquad\CK:=\Sw(X)^K,\qquad\phiz:=\one_{X(\cO)}\in\CK .\]\(\phiz\) is the . It depends on the model of \(X\) over \(\cO\), not only on \(X\) over \(F\).
On the running examples (K-orbits): \(\phiz=\one_{\cO^\times}\) for \(\GL_2/\SL_2\); \(\phiz=\one_{\{d=0\}}\) for \(\GL_2/T\); \(\phiz=\one_{\GL_2(\cO)}\), the unit of the Hecke algebra of \(\GL_2\), for the group; \(\phiz=\one_{\cO^n}\) for \(\Aff^n\).
The spherical Hecke algebraWhat acts on \(\CK\)?
[Cartier, §1.3–1.4] Definition
With Haar measure \(\vol K=1\), \(\HK:=C_c(K\backslash G(F)/K)\) acts on \(\CK\) by
\[(f\ast\varphi)(x):=\int_{G(F)}f(g)\,\varphi(g^{-1}x)\,dg=\sum_{g\in G(F)/K}f(g)\,\varphi(g^{-1}x).\]By the Cartan decomposition \(G(F)=\bigsqcup_{\lambda\in\LTp}Kt^\lambda K\), \(t^\lambda:=\lambda(\vpi)\), the operators \(T_\lambda:=\one_{Kt^\lambda K}\) form a basis of \(\HK\).
The summand depends only on \(gK\); finitely many cosets contribute; \(f\ast\varphi\) is again \(K\)-invariant and compactly supported. The same formula makes \(\HK\) an associative algebra with unit \(\one_K\), commutative by Satake (next page). For \(\GL_n\) write \(T_1:=T_{(1,0,\dots,0)}\) and \(Z:=T_{(1,\dots,1)}=\one_{\vpi K}\), which is central and invertible.
(Hecke translates of \(\phiz\)) Proved here
\[(T_\lambda\ast\phiz)(x)=\#\{gK\subseteq Kt^\lambda K:\ x\in g\,X(\cO)\}\in\Z_{\ge0},\qquad\supp(T_\lambda\ast\phiz)=Kt^\lambda\,X(\cO).\]Proof
With \(\vol K=1\), \((T_\lambda\ast\phiz)(x)=\sum_{gK\subseteq Kt^\lambda K}\one_{X(\cO)}(g^{-1}x)\), and the term for \(gK\) is \(1\) iff \(x\in gX(\cO)\). So the support is \(Kt^\lambda K\cdot X(\cO)=Kt^\lambda X(\cO)\), as \(X(\cO)\) is \(K\)-stable.
For \(\Aff^n\), \(gX(\cO)=g\cO^n\) is a lattice, and \(T_\lambda\ast\phiz(v)\) counts lattices of relative position \(\lambda\) that contain \(v\).
Spherical generationIs every \(K\)-invariant Schwartz function a Hecke translate of \(\phiz\)?
Definition
\[\boxed{\ \text{(SG)}\qquad\HK\ast\phiz=\CK\ }\]For homogeneous \(X\), \(\CK\) has the basis of orbit indicators (K-orbits), so (SG) says: every \(\one_{Kx}\) is some \(f\ast\phiz\).
A triangular criterionHow does one prove (SG) by hand?
(triangular generation) Proved here
Let \(\Omega\) be a set of compact open \(K\)-orbits whose indicators span \(\CK\), with a partial order in which every lower set \(\{\nu\le\mu\}\) is finite. If every \(\mu\in\Omega\) has some \(f_\mu\in\HK\) with
\[\supp(f_\mu\ast\phiz)\subseteq\bigcup_{\nu\le\mu}O_\nu\qquad\text{and}\qquad f_\mu\ast\phiz\ne0\ \text{on}\ O_\mu,\]then (SG) holds. Conversely, (SG) gives such \(f_\mu\) for every order, with \(f_\mu\ast\phiz=\one_{O_\mu}\).
Proof
Induction on \(|\{\nu\le\mu\}|\). \(f_\mu\ast\phiz\) is constant on orbits, so \(f_\mu\ast\phiz=c_\mu\one_{O_\mu}+\sum_{\nu\lt\mu}c_\nu\one_{O_\nu}\) with \(c_\mu\ne0\). By induction every \(\one_{O_\nu}\), \(\nu\lt\mu\), lies in \(\HK\ast\phiz\); hence so does \(\one_{O_\mu}\).
(\(\GL_2/T\) on the tree)
\(T_1\) is the adjacency operator of the tree, so on the orbits \(d\) of the orbit page:
\[T_1\one_{\{0\}}=2\,\one_{\{0\}}+\one_{\{1\}},\qquad T_1\one_{\{1\}}=(q-1)\one_{\{0\}}+\one_{\{2\}},\qquad T_1\one_{\{d\}}=q\,\one_{\{d-1\}}+\one_{\{d+1\}}\ (d\ge2).\]Order by \(d\) and take \(f_d=T_1^d\): (SG) holds. The central \(Z\) acts trivially.
How the three formulas are computed
\((T_1\one_S)(x)\) is the number of neighbours of \(x\) in \(S\) (identify \(K\backslash X(F)\) with the \(T(F)\)-orbits of vertices, the distance to the apartment [Serre, Trees, II.1]). An apartment vertex has \(2\) neighbours on the apartment and \(q-1\) off it; a vertex at distance \(d\ge1\) has one neighbour at \(d-1\) and \(q\) at \(d+1\).
The Hecke labCan you reach a single orbit from \(\phiz\)?
The lab stores \(f\) together with a Hecke word \(h\) in \(T_1,Z^{\pm1}\) with \(f=h\ast\phiz\). It computes the spectral image twice: \(\Phi(f)\) from the values of \(f\) on orbits, and \(\Res(h)\) from the word, by the formulas of the Theorem page (tree), A* and A_X* (\(\Gm\)) and 𝔸ⁿ. Their agreement on every input is a computed check of those formulas, not a proof. Challenge on the tree: reach \(\one_{\{d=3\}}\); the solution is the Proposition above. On \(\Aff^n\): reach \(\one_{X_0}\) (save \(Z\ast\phiz\), go back with \(Z^{-1}\), subtract).
The group caseWhat is (SG) when \(X\) is a group?
(group case) Proved here
For \(X=H\) under \(H\times H\), \(\CK=\mathcal H_{K_H}\), \(\phiz=\one_{K_H}\) is its unit, \(a\otimes b\) acts by \(\varphi\mapsto a\ast\varphi\ast b\), and
\[\HK\ast\phiz=\mathcal H_{K_H}\ast\one_{K_H}\ast\mathcal H_{K_H}=\mathcal H_{K_H}=\CK .\]In the triangular language: the orbits are the double cosets \(K_Ht^\lambda K_H\), and \(f_\lambda=T_\lambda\otimes1\) hits exactly one orbit; the system is diagonal.