A question on non-archimedian Fourier transform

Question

Let $M(n)$ be the vector space of $n\times n$ matrices over a local non-archimedian field $K$. Let $\mathcal S$ denote the space of locally constant compactly supported functions on $M(n)$. Similarly, let $\mathcal S ^0$ denote the space of locally constant compactly supported functions on the group $GL(n)$ (which is an open subset of $M(n)$).

Let us identify $M(n)$ with its dual space by means of the pairing $(x,y)=Tr(xy)$. Let $F$ denote the corresponding Fourier transform acting from $\mathcal S$ to itself (formally, $F$ depends on a choice of a non-trivial additive character of $K$, but it does not matter for the question). Is this true that $$ \mathcal S=\mathcal S^0 + F(\mathcal S^0)? $$ This is obvious when $n=1$, but already in the case $n=2$ the asnwer is not clear to me.

Answer 1

Can one try to show the negative answer for n=2 as follows. The question is equivalent to existence of a nonzero distribution σ on a 4 dimensional space $K^4$ such that both both σ and FT(σ) are supported on the quadric ab=cd (where a,b,c,d are coordinates on $K^4$). For a quadric in $K^3$ (a.k.a. the nilpotent cone in sl(2)) there is a standard way to do it: there is a unique up to scaling distribution supported there which is invariant under conjugations by SL(2) but transforms under a non-trivial character under conjugations by GL(2). Can one do something like this here replacing GL(2) by Spin(4)?

Answer 2

Actually, the answer is indeed negative and the explanation is very simple. Namely, for $n=2$ let $\phi$ be the delta-function of the space of matrices whose second row is zero (considered as a distribution). Then its Fourier transform is the delta-function of matrices whose first column is zero. Hence this is a distribution with support on degenerate matrices whose Fourier transform is also concentrated on degenerate matrices.

Answer 3

Sorry but I would like to change my vote and would now argue for a positive answer. I am afraid I made a mistake when saying that I know a distribution on $sl(2)$ supported on the nilcone whose FT also has this property. I know one for $sl(2)$ over a finite field but not over a local non-Archimedian field. In this latter context it is known (at least when characteristic of the local field is zero) that FT of a nilpotent orbit comes from an $L^1$ function on the set of regular semi-simple elements, this shows that FT of a nilpotently supported invariant distribution can not be nilpotently supported. Another argument that might show the same without assuming the distribution is invariant is like this: the metaplectic group Mp(2) acts on the space of distributions where upper triangular matrices with 1 on diagonal act by multiplication by an additive character of the quadratic form and the order 4 element acts by FT. If both $\psi$ and FT of $\psi$ are supported on the nilcone, then $\psi$ is invariant under this action. Then perhaps the center of Mp(2) acts by a nontrivial character, so no nonzero distribution is invariant under Mp(2)?

Answer 4

Belatedly, but perhaps of some interest, after some intermittent thought: for non-archimedean $k$ of characteristic $0$ (maybe un-necessarily), non-archimedean, let $A$ be the $n$-by-$n$ matrices, $A^{\le r}$ the matrices of rank $\le r$, $A^{\ge r}$ those of rank at least $r$, and $A^r$ those of rank exactly $r$. Let $G=GL_n(k)$ and $G^1=\{g\in G:|\det g|=1\}$. Let $K=GL_n(\mathfrak o)$, where $\mathfrak o$ is the local integer ring.

We can ask for left-and-right $G^1$-invariant distributions supported on $A^{\le r}$ with $r<n$. Interestingly, $K\times G^1$ is transitive on $A^r$ under the action $(A,B)(x)=A^{-1}xB$. Thus, by uniqueness of invariant distributions, the restriction to $A^{\ge r}$ of a $K\times G^1$-invariant, or $G^1\times K$-invariant distribution supported on $A^\le r$, is unique up to scalar multiples. Indeed, a $G^1\times G^1$-invariant distribution is unique.

In particular, for Schwartz function $f$, the following three integrals give the same outcome up to constants: $$ \int_K \int_{r\times n} f(k\cdot \pmatrix{0_{n-r} & 0 \cr x_{21} & x_{22}})\;dx_{21}\,dx_{22}\,dk $$ $$ \int_K \int_{n \times r} f(\pmatrix{0_{n-r}&x_{12}\cr 0 & x_{22}}\cdot k)\;dx_{12}\,dx_{22}\,dk $$ $$ \int_{G^1\times G^1/\Theta} f(g^{-1}\cdot \pmatrix{0_{n-r} & 0 \cr 0 & 1_r}\cdot g')\;dg\,dg' $$ where $\Theta$ is the isotropy group of $\pmatrix{0_{n-r} & 0 \cr 0 & 1_r}$. Miraculously, $\Theta$ is unimodular, so there is indeed an invariant measure on that quotient.

In particular, the $G\times G$-equivariance is determined by $r$, from those integral formulas: the distribution $v_r$ attached to $r<n$ is homogeneous of degree $r$, in the sense that $v_r(x\to f(A^{-1}xB))=|\det A|^r \cdot |\det B|^r\cdot v_r(f)$: the left equivariance follows from the second integral expression, the right equivariance from the first integral expression.

Fourier transform sends degree-$r$-homogeneous distributions to degree $n-r$. For $n=1$, there is no room to have a $G^1\times G^1$-invariant distribution with support on singular matrices whose Fourier transform is supported likewise. But for $n\ge 2$, there are unique candidates for those invariant distributions, with a unique (up to scalars) distribution for each $r$ in the range $0<r<n$, and Fourier transform replacing $r$ by $n-r$ (so rank $0$ support has Fourier transform consisting of constants, with non-singular support).

The meromorphic family of equivariant distributions $u_s=|\det x|^{s-n}$ has a meromorphic continuation, and the $s$-th one is homogeneous of degree $s$. The residue at $s=r$ for $0<r<n$ is a constant multiple of $v_r$...