Timeline for A question on non-archimedian Fourier transform
Current License: CC BY-SA 3.0
16 events
| when toggle format | what | by | license | comment | |
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| Aug 15, 2014 at 3:32 | comment | added | asv | Roman, I do not see how to modify my example to $sl(2)$. | |
| Aug 14, 2014 at 19:36 | comment | added | Roman | 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. | |
| Aug 14, 2014 at 18:39 | comment | added | Alexander Braverman | By nilcone he actually means degenerate matrices, but there is one point in his proof that I don't understand. | |
| Aug 14, 2014 at 18:08 | comment | added | asv | 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. | |
| Aug 14, 2014 at 17:14 | comment | added | Alexander Braverman | OK, I agree (about the invariance of the condition - I just made a stupid computational mistake). Well, in this case we seem to have a contradition in mathematics, since I also don't see any flaw in Roman's argument (which shows that such a distribution can't exist...). | |
| Aug 14, 2014 at 16:56 | comment | added | asv | Sasha, I think the equation is invariant under the Fourier transform; this is straightforward computation. If you wish I can write it down. The proof of uniqueness of the distribution satisfying the above equation is known to me in the real case. In the non-archimedian case there are some technical difficulties which I did not think through (like restriction of an equivariant distribution to a transversal to an orbit). | |
| Aug 14, 2014 at 15:19 | comment | added | Alexander Braverman | Actually, I think that the equation $\psi(AXB)=|\det(A)\det(B)|^{-1}\psi(X)$ is not invariant under the Fourier transform. Am I wrong? | |
| Aug 14, 2014 at 15:06 | comment | added | Alexander Braverman | Yes, sure, I got confused too about something. Semyon, why is this distribution unique? | |
| Aug 13, 2014 at 2:06 | comment | added | Roman | 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. | |
| Aug 12, 2014 at 23:14 | comment | added | Alexander Braverman | Semyon, I need compacts support but in the non-archimedian case there should be no difference. Let me try to understand your distribution. | |
| Aug 12, 2014 at 16:45 | comment | added | David E Speyer | Never mind, I was confused. The point is that $\mathcal{S}^0$ and $F(\mathcal{S}^0)$ should be contained in $\psi^{\perp}$; we don't need $\psi$ itself to be compactly supported. | |
| Aug 12, 2014 at 14:29 | comment | added | asv | @DavidSpeyer: this is true that $\psi$ is not compactly supported. However it is not supported on nilpotent cone, but rather on matrices of rank less than 2. $\psi$ is a distribution of tempered growth; also it does not die off expotentially at infinity. | |
| Aug 12, 2014 at 14:17 | comment | added | David E Speyer | @semyonalesker The equation $\psi(AXB) = |\det(A) \det(B)|^{-1} \psi(X)$ forces $\psi$ to be supported on the entire nilpotent cone, and thus not compactly supported. It looks like it might die off exponentially fast as we get away from $0$, though. Prof. Braverman, do you really need compact support, or is rapid decay good enough? | |
| Aug 12, 2014 at 14:08 | comment | added | asv | 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. | |
| Aug 12, 2014 at 13:26 | comment | added | asv | 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 $\mathcal{S}$. | |
| Aug 12, 2014 at 13:08 | history | answered | Roman | CC BY-SA 3.0 |