Computations with Bernstein Projectors of SL(2)

Abstract

For the p-adic group G = SL(2), we present results of the computations of the sums of the Bernstein projectors of a given depth. Motivation for the computations is based on a conversation with Roger Howe in August 2013. The computations are elementary, but they provide an expansion of the delta distribution $\delta _{1_{G}}$ into an infinite sum of G-invariant locally integrable essentially compact distributions supported on the set of topologically unipotent elements. When these distributions are transferred, by the exponential map, to the Lie algebra, they give G-invariant distributions supported on the set of topologically nilpotent elements, whose Fourier transforms turn out to be characteristic functions of very natural G-domains. The computations in particular rely on the SL(2) discrete series character tables computed by Sally–Shalika in 1968. This new phenomenon for general rank has also been independently noticed in recent work of Bezrukavnikov, Kazhdan, and Varshavsky.

In honor of Roger Howe as a septuagenarian, andin memory of Paul Sally Jr. and Joseph Shalika

The author is partly supported by Hong Kong Research Grants Council grant CERG #603813.

Appendix A

Here G = SL(2, F), and $\mathfrak{g}\, =\, \mathfrak{s}\mathfrak{l}(2,F)$. Suppose ψ is an additive character of F with conductor ℘. Let FT denote the Fourier transform on $\mathfrak{g}$, i.e., if $f \in C_{c}^{\infty }(\mathfrak{g})$:

$$\displaystyle{\text{FT}\,(f)(Y )\ =\ \int _{\mathfrak{g}}\psi \,(\,\text{trace}(\,X\,Y \,)\,)\ f(X)\ dX.}$$

In this appendix we prove:

Proposition 1

For $\mathfrak{g}\, =\, \mathfrak{s}\mathfrak{l}(2,F)$ , we have

• The Fourier transforms $FT(1_{\mathfrak{g}_{0}})$ and $FT(1_{\mathfrak{g}_{ -\frac{1} {2} }})$ have support in the sets $\mathfrak{g}_{0^{+}}:= \mathfrak{g}_{\frac{1} {2} }$ and $\mathfrak{g}_{(\frac{1} {2} )^{+}}:= \mathfrak{g}_{1}$ , respectively. In particular, the support is contained in $\mathcal{N}^{\mathit{\text{top}}}$.

• For k ≥ 1, the Fourier transform $FT(1_{\mathfrak{g}_{-k}})$ has support in $\mathfrak{g}_{k^{+}}:= \mathfrak{g}_{k+\frac{1} {2} }$.

Proof

Since $1_{\mathfrak{g}_{0}}$, and $1_{\mathfrak{g}_{ -\frac{1} {2} }}$ are Ad (G)-invariant sets, their Fourier transforms are Ad (G)-invariant. Therefore, it is sufficient to show the stated vanishing on any convenient element in an Adjoint orbit.

We prove the result for $1_{\mathfrak{g}_{0}}$ and remark our argument proof is easily adapted to also treat the case $1_{\mathfrak{g}_{ -\frac{1} {2} }}$. We have

$$\displaystyle{\text{FT}\,(1_{\mathfrak{g}_{0}})\,(\,Y \,)\ =\ \text{PV}\,\int _{\mathfrak{g}}\ \psi (\,\text{trace}(\,X\,Y \,)\ 1_{\mathfrak{g}_{0}}(\,X\,)\ dX\quad (\text{principal value}).}$$

We note that $\mathfrak{g}_{0}\, =\,\{\, X \in \mathfrak{g}\ \vert \ \det (X) \in \mathcal{R}_{F}\,\}$. Let $X\, =\, \left [\begin{array}{*{10}c} a& b\\ c &-a \end{array} \right ]$, and for integral ℓ, set

$$\displaystyle{\mathcal{T}_{\ell}\ =\ \{\ X \in \mathfrak{g}_{0}\ \vert \ a,\,b,\,c\ \in \wp ^{-\ell}\ \}\ =\ \{\ X \in \mathfrak{g}\ \vert \ \det (X) \in \mathcal{R}_{ F}\ \text{and}\ a,\,b,\,c\ \in \wp ^{\ell}\ \}.}$$

We show for $Y \not\in \mathfrak{g}_{\frac{1} {2} }$ the integral

$$\displaystyle{\int _{\mathcal{T}_{\ell}}\ \psi (\,\text{trace}(\,X\,Y \,)\,)\ 1_{\mathfrak{g}_{0}}(\,X\,)\ dX\ =\ \int _{\mathcal{T}_{\ell}}\ \psi (\,\text{trace}(\,X\,Y \,)\,)\ dX\quad \text{vanishes for }\ell\text{ large.}}$$

The Fourier transforms $\text{FT}(1_{\mathfrak{g}_{j}})$ are invariant under the Adjoint action of GL(2, F). The GL(2, F)-orbit of a regular semisimple element Y contains an element of the form

$$\displaystyle{\left [\begin{array}{*{10}c} 0 &B\\ C & 0 \end{array} \right ]\quad \text{with val}(B)\, \leq \,\text{val}(C)\, \leq \,\text{val }(B) + 1.}$$

We take Y to have this anti-diagonal form. Here, Y is not topologically nilpotent when val(C) ≤ 0. It thus suffices to prove $\text{FT}(1_{\mathfrak{g}_{0}})(Y )\, =\, 0$ in this situation. For $X\, =\, \left [\begin{array}{*{10}c} a& b\\ c &-a \end{array} \right ]$, we have trace( XY ) = (bC + cB), and so

$$\displaystyle{\int _{\mathcal{T}_{\ell}}\ \psi (\,\text{trace}(\,X\,Y \,)\,)\ dX\ =\ \int _{\mathcal{T}_{\ell}}\ \psi (\,bC\, +\, cB\,)\ dX.}$$

CASE val(C) = val(B): We remark in this situation, Y is either split or elliptic unramified. We show the integral vanishes if B, C ∉ ℘. Our strategy is to partition $\mathcal{T}_{\ell}$ into regions where the integral is zero.

SUBCASE $a \in \mathcal{R}_{F}$ : The condition for X to be in $\mathfrak{g}_{0}$ is $a^{2} + bc \in \mathcal{R}_{F}$, and so $bc \in \mathcal{R}_{F}$. We consider two subcases based on b.

• Subcase $b \in \mathcal{R}$. Here, depending on the valuation val(b), the variable c runs over an ideal between ℘ ⁰ and ℘ ^−ℓ. The assumption B ∉ ℘ means c → ψ(cB) is a non-trivial character on its allowed ideal and therefore, for fixed a and b, the integral over c is zero. We deduce the integral over the region in $\mathcal{T}_{\ell}$ which satisfies $a,b \in \mathcal{R}$ is zero.

• Subcase b = ϖ ^−k u with (integral) k > 0 and u a unit. The condition $bc \in \mathcal{R}$, is c ∈ ℘ ^k ⊂ ℘. If val(B) ≤ −k, then c → ψ(cB) is a non-trivial character; so integration of c over ℘ ^k is zero. If − k < val(B), then ψ(cB) = 1; so ψ(bC + cB) = ψ(bC). Integration over c ∈ ℘ ^k yields ψ(bC) meas(℘ ^k). We note that x → ψ(xC) is a non-trivial character on $\mathcal{R}$, and ψ((b + x)C) = ψ(bC)ψ(xC). If we integrate over all b ∈ ℘ ^−k∖℘ ^−k+1, we get zero. We deduce the integral over the region in $\mathcal{T}_{\ell}$ which satisfies $a \in \mathcal{R}$ and $b\not\in \mathcal{R}$ is zero.

CASE $a\not\in \mathcal{R}_{F}$ : Write a as a = ϖ ^−k u with u a unit, and k a positive integer (note k ≤ ℓ to satisfy a ∈ ℘ ^−ℓ). The condition $a^{2}\, +\, bc\, \in \,\mathcal{R}$ for X to be in $\mathfrak{g}_{0}$ is thus $-bc\, \in \,\varpi ^{-2k}u^{2}\, +\, \mathcal{R}$. In particular, b and c are non-zero. Write b, as b = ϖ ^β v _b, with v _b a unit. The condition, b, c ∈ ℘ ^−ℓ means − ℓ ≤ β ≤ ℓ − 2k, and similarly for the valuation γ = val(c) of c.

• If β ≠ γ, then by the symmetry of b and c, we assume β < −k < γ. This imposes the condition c ∈ −ϖ ^γ v _b ⁻¹ u ² + ℘ ^−β.

• If val(B) −β ≤ 0, then for fixed a and b, the integral over c is zero.

• If val(B) −β > 0, then ψ(cB) = ψ(−ϖ ^γ v _b ⁻¹ u ² B) is independent of c ∈ −ϖ ^γ v _b ⁻¹ u ² + ℘ ^−β. Thus, if we fix a and b, and integrate ψ(bC + cB) over c, we get

  $$\displaystyle{\psi (bC)\,\psi (-\varpi ^{\gamma }\,v_{b}^{-1}\,u^{2}\,B)\,\text{meas}(\varpi ^{(-2k-\beta )} + \wp ^{-\beta }).}$$

  If we perturb b by v _b x ∈ ℘ ^−val(B) to b ^′ = b + v _b x, so $v_{b^{{\prime}}} = v_{b}(1 +\varpi ^{-\beta }x)$, then the corresponding c ^′ satisfies c ^′ ∈ −ϖ ^γ v _b ⁻¹(1 + ϖ ^−β x)⁻¹ u ² + ℘ ^−β = −ϖ ^γ v _b ⁻¹ u ² + ϖ ^γ−β v _b ⁻¹ xu ² + ℘ ^−β. We deduce ψ(c ^′ B) = ψ(−ϖ ^γ v _b ⁻¹ u ² B) is independent of x. So, ψ(b ^′ C + c ^′ B) = ψ(bC) ψ(xC) ψ(−ϖ ^γ v _b ⁻¹ u ² B), and therefore, if we restrict to $b \in \varpi ^{-\beta }\mathcal{R}^{\times }$ integrate over c ^′ followed by integration over b, we get zero.

• If β = γ = −k, we have

  $$\displaystyle{X\ =\ \varpi ^{-k}\left [\begin{array}{*{10}c} u & v_{b} \\ v_{c}&-u\end{array} \right ]\quad \text{with }u,\,v_{b},\,v_{c}\text{ units},}$$

  and the condition for $X \in \mathfrak{g}_{0}$ is u ² + v _b v _c ∈ ℘ ^2k. We see the product − v _b v _c must be a square in $\mathcal{R}^{\times }$. Conversely, if − v _b v _c is a square, the condition on u is u ² ∈ −v _b v _c + ℘ ^2k. We fix v _c and multiplicatively perturb v _b by 1 + ℘, the integral of $\psi (\varpi ^{-k}(v_{c}B + v_{b^{{\prime}}}C))$ over b ^′ ∈ ϖ ^−k v _b(1 + ℘), and a = ϖ ^−k u with $u^{2}\, \in \,-v_{b^{{\prime}}}v_{c} + \wp ^{2k}$. The integration over a yields a constant independent of b ^′, and then the integration over b ^′ is zero. So the integral of $\psi (\varpi ^{-k}(v_{c}B + v_{b^{{\prime}}}C))$ over the region satisfying $X \in \mathfrak{g}_{0}$ and $a,\,b,\,c\, \in \varpi ^{-k}\mathcal{R}^{\times }$ is zero.

CASE val(C) = val(B) + 1: The proof here is a minor modification of the case val(C) = val(B). We omit the details.

This completes the proof that $\text{FT}(1_{\mathfrak{g}_{0}})$ has support in $\mathfrak{g}_{\frac{1} {2} }$.

The statement about the support of $\text{FT}(1_{\mathfrak{g}_{-k}})$ for k ≥ 0 follows from the elementary property $\mathfrak{g}_{k-1}\, =\,\varpi ^{-1}\mathfrak{g}_{k}$, and elementary homogeneities of the Fourier transform. □

Appendix B