Two bases, one generator

What is Proposition III saying about locally constant Iwahori-invariant functions?

1 · Two vector-space bases at a finite level

Use the lattices of the model, and define functions $\basisLat_{\idx},\basisShell_{\idx}:\Loc^{\nDim}\to\mathbb C$:

$$\basisLat_{\idx}=\mathbf1_{\Lat_{\idx}},\qquad\basisShell_{\idx}=\mathbf1_{\Lat_{\idx}\setminus\Lat_{\idx+1}}.$$

For integers $\lo\le\hi$, let $\Window_{\lo,\hi}$ be the I-invariant functions supported in $\Lat_{\lo}$ and constant on cosets of $\Lat_{\hi}$. The finite quotient $\Lat_{\lo}/\Lat_{\hi}$ has strata $\Lat_{\idx}\setminus\Lat_{\idx+1}$ for $\lo\le\idx<\hi$, together with its zero stratum.

· Shells and nested supports Proved here

$$\mathcal B_{\mathrm{std}}=\{\basisShell_{\lo},\ldots,\basisShell_{\hi-1},\basisLat_{\hi}\}.$$ $$\mathcal B_{\mathrm{cl}}=\{\basisLat_{\lo},\ldots,\basisLat_{\hi}\}.$$

Both are bases of $\Window_{\lo,\hi}$, and the change of basis is triangular:

$$\basisShell_{\idx}=\basisLat_{\idx}-\basisLat_{\idx+1}\quad(\lo\le\idx<\hi).$$ $$\basisLat_{\idx}=\sum_{\idx\le j<\hi}\basisShell_j+\basisLat_{\hi}.$$

Proof

Every nonzero vector lies in a unique shell. In each shell, elementary upper-triangular integral operations carry a vector to its leading coordinate vector; the finite quotient has the same orbit list. Thus invariance means being constant on each listed stratum, giving $$\dim\Window_{\lo,\hi}=\hi-\lo+1.$$
The detailed elimination is the orbit lemma.
The shells and the terminal lattice partition the support: $$\Lat_{\idx}=\bigsqcup_{\idx\le j<\hi}(\Lat_j\setminus\Lat_{j+1})\ \sqcup\ \Lat_{\hi}.$$
The resulting triangular matrix has diagonal entries one: $$\det(U)=1.$$

· A four-stratum computation Example

Take $\nDim=2$, $\resq=3$, and indices $\lo=0$, $\hi=3$. In $\Lat_0/\Lat_3$, the three shell sizes and terminal stratum size are

A function with successive values $(2,5,5,7)$ has the two expansions

$$f=2\basisShell_0+5\basisShell_1+5\basisShell_2+7\basisLat_3.$$ $$f=2\basisLat_0+3\basisLat_1+0\basisLat_2+2\basisLat_3.$$

Proof

$$|\Lat_j/\Lat_3|=3^{3-j}\quad(0\le j\le3).$$$$(2,5,5,7)\longmapsto(2,5-2,5-5,7-5)=(2,3,0,2).$$

2 · One module generator

· The function-theoretic conclusion Proved here

Define $\basic=\basisLat_0$ and $\Module=C_c^\infty(\Loc^{\nDim})^{\Iw}$. Then

$$T_{\shift^{\idx}}*\basic=\basisLat_{\idx}.$$ $$\boxed{\mathbb C[\Om]*\basic=\Module=\Hecke*\basic.}$$

In fact the linear map $\mathbb C[z,z^{-1}]\to\Module$, $z^{\idx}\mapsto\basisLat_{\idx}$, is an isomorphism of $\mathbb C[\Om]$-modules.

Proof

The normalizer calculation gives $$\basic(\shift^{-\idx}x)=\mathbf1_{\shift^{\idx}\Lat_0}(x)=\basisLat_{\idx}(x).$$
Compact support bounds a function inside some $\Lat_{\lo}$; local constancy and compactness give a common sufficiently small lattice $\Lat_{\hi}$ of translation invariance. Therefore $$\Module=\bigcup_{\lo\le\hi}\Window_{\lo,\hi}.$$
Use the finite-level basis result for spanning. For independence, evaluate a finite linear combination successively on its outermost shells: $$\sum_{\idx=\lo}^{\hi}c_{\idx}\basisLat_{\idx}=0\ \Longrightarrow\ c_{\lo}=c_{\lo+1}=\cdots=c_{\hi}=0.$$

The two families are vector-space bases. The vector $\basic$ is one module generator; the operator $T_{\shift}$ and its inverse move it through the lattice basis.

3 · The zero point and the IC basis

At a finite level the final basis vector is $\basisLat_{\hi}$, the lift of the zero-orbit indicator. On the full local field, $\mathbf1_{\{0\}}$ is not locally constant.

$$\operatorname{span}_{\mathbb C}\{\basisShell_{\idx}:\idx\in\mathbb Z\}=\ker\bigl(\mathrm{ev}_0:\Module\to\mathbb C\bigr).$$

Thus bare shell indicators are not a basis of all $\Module$; the terminal lattice cannot be discarded in a finite expansion. The full space also has the basis $\{\basic\}\cup\{\basisShell_{\idx}:\idx\in\mathbb Z\}$.

For the geometric category over $\overline{\mathbb F}_{\resq}$, with the dimension convention of Proposition III, put $\Simple_{\idx}=\underline{\overline{\mathbb Q}}_{\ell,\Lat_{\idx}}[-\idx]$. The trace functions satisfy

$$\operatorname{tr}(\Simple_{\idx})=(-1)^{\idx}\basisLat_{\idx}.$$

All closures are smooth linear subspaces, so the IC basis is the closure-indicator basis up to these signs. The general two-basis comparison in Chen–Yi concerns symmetric varieties; here the displayed finite triangular calculation proves the affine-space instance.

[CY, §8.2.3 and Lemma 73, pp. 42–43]

Which parts of this calculation are sufficient in general? →