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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.

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    $\begingroup$ I would ordinarily think of a map $\mathcal S\to \mathcal S^0$, but you're regarding $\mathcal S^0$ as a subspace of $\mathcal S$. Is this by taking functions on $M(n)$ supported inside $GL(n)$, or is it by using the inner product somehow? $\endgroup$ – Allen Knutson Aug 11 '14 at 19:36
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    $\begingroup$ No, there is only a map from $\mathcal S^0$ to $\mathcal S$ (since both are spaces of compactly supported functions, there is no map in the opposite direction). You can think of $\mathcal S^0$ as the subspace of $\mathcal S$ consisting of functions which vanish on degenerate matrices. $\endgroup$ – Alexander Braverman Aug 12 '14 at 6:53
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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)?

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    $\begingroup$ I think this answer does work in the real case (and hopefully in the non-archimedian one too). There is a unique distribution $\psi$ on $M(2)$ which satisfies $\psi(AXB)=|\det(A)\det(B)|^{-1}\psi(X)$ for any $A,B\in GL(2),X\in M(2)$. Moreover this $\psi$ is supported on degenerate matrices. Since the former condition is stable under Fourier transform, $F(\psi)$ is supported on that set too by uniqueness. Hence $\psi$ vanishes on $\mathcal{S}^0+F(\mathcal{S}^0)$, and the latter set cannot be dense in $\mathcal{S}$. $\endgroup$ – MKO Aug 12 '14 at 13:26
  • $\begingroup$ The above mentioned distribution $\psi$ can be obtained e.g. by meromorphic continuation of the function $|\det|^s$ to the point $s=-1$ and taking the coefficient of the highest pole. In the archimedian case it can be done using Bernstein's theorem on b-function. In non-archimedian case this meromorphic continuation is known to exit in the case of zero characteristic. In the positive characteristic case I do not know if such existence is proven in general. $\endgroup$ – MKO Aug 12 '14 at 14:08
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    $\begingroup$ Sasha, you don't need compact support of distribution $\psi$ (I called is $\sigma$). Existence of such a distribution, whatever its support implies a negative answer to your question, as the your space of functions will be contained in the orthogonal to that distribution. $\endgroup$ – Roman Aug 13 '14 at 2:06
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    $\begingroup$ I do not see the contradiction. I claim that there is a distribution supported on degenerate matrices in $M(2)$ such that its Fourier transform has the same property. Roman claims that there is no distribution supported on the nil-cone of $sl(2)$ such that its Fourier transform has the same property. $\endgroup$ – MKO Aug 14 '14 at 18:08
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    $\begingroup$ Semyon, isn't there a modification of your example which would apply to $sl(2)$ producing a Fourier invariant (up to a scalar) distribution there supported on the nilcone? If not, then the contradiction in math is avoided, but there are two curious points: a psychological one, that my original answer was correct, though my understanding of the motivating "baby case" was mistaken; and a mathematical one, that some naive generalizations of the $L^1$-property of FT of invariant distributions supported on the nilcone are false. $\endgroup$ – Roman Aug 14 '14 at 19:36
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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.

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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)?

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  • $\begingroup$ I didn't understand the last argument - on what space does the group Mp(2) act? $\endgroup$ – Alexander Braverman Aug 14 '14 at 15:02
  • $\begingroup$ Below Semyon claims that there is a distribution on degenerate matrices whose FT is concentrated there as well, but so far I don't understand all the details. $\endgroup$ – Alexander Braverman Aug 14 '14 at 15:09
  • $\begingroup$ Mp(2) acts on functions and hence on distributions on $K^4=gl(2)$ (as well as $K^3=sl(2)$). In fact a bigger group Mp(8) (resp. Mp(6)) acts there (and its central element surely acts but a nontrivial scalar), then the quadratic form $det$ defines a homomorphism $Mp(2)\to Mp(8)$. $\endgroup$ – Roman Aug 14 '14 at 16:04
  • $\begingroup$ Actually I don't understand why the center of $Mp(2)$ acts non-trivially in this representation. $\endgroup$ – Alexander Braverman Aug 14 '14 at 17:29
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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 $v_r$...

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