Let $\mathcal{P}=\{\infty, 2,3,5,7,11,\ldots\}$ be the set of primes of $\mathbb{Q}$ and let $\mathbb{Q}_p$ denote the corresponding completions, so in particular $\mathbb{Q}_{\infty}=\mathbb{R}$.

Is it possible to give a definition of the spaces of test functions $S(\mathbb{Q}_p)$ together with their standard topologies (finest locally convex for $p$ finite and usual Frechet in the real case) in a completely uniform way with respect to $p\in\mathcal{P}$?

It is not too hard to uniformly define $L^2(\mathbb{Q}_p, dx)$ as well as the Fourier transform as a unitary transformation on this space. The kind of definition I am looking for might look like: $S(\mathbb{Q}_p)$ is the smallest subspace of $L^2$ which is invariant by Fourier transform and... insert mystery property here...

  • $\begingroup$ I Guess because you are using the letter S you mean the schwartz space rather than test functions... In this case I would say "the largest subspace of $L^2$ stable under Fourier transform and multiplication by |x|". Also, from this perspective, the definition of the topology in the real case carry over to the p-adic space, I'm not exactly sure if it is the correct topology... $\endgroup$ – Simon Henry Sep 7 '15 at 15:39
  • $\begingroup$ How does that force local constancy in the p-adic case? $\endgroup$ – Abdelmalek Abdesselam Sep 7 '15 at 16:56
  • $\begingroup$ You're right it does not work, but the following do: It is the largest subspace of $L^2$ stable under fourier transform, multiplication by $|x|$ and multiplication by locally constant functions. $\endgroup$ – Simon Henry Sep 8 '15 at 8:53
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    $\begingroup$ I would like like answer where smoothness in the real case and local constancy in the p-adic case emerge a posteriori instead of being put in by hand in the definition. $\endgroup$ – Abdelmalek Abdesselam Sep 10 '15 at 14:39
  • $\begingroup$ Well, in the definition I gave you smoothness and local constancy do emerge a posteriori: the fact that you want something which is stable by multiplication by locally constant function does not implies directly that the function in the space should be them selve locally constant. But here is the problem: in the real case smoothness corresponds to quickly decreasing Fourier transform, while in the p-adique case local constancy corresponds to compactly supported Fourier transform, and unfortunately it is possible to define a notion of smoothness in the p-adic case: one can say... $\endgroup$ – Simon Henry Sep 10 '15 at 14:53

François Bruhat in his paper of 1961 (see page 61) gave a definition (and proved some properties) of the space ${\mathcal S}(G)$ of the Schwartz test functions on an arbitrary abelian locally compact group $G$.

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    $\begingroup$ yes but I find it a bit a cheat. As far as I know, Bruhat's description uses the rather heavy stucture theory of ALC groups and differentiability is put in by hand. $\endgroup$ – Abdelmalek Abdesselam Sep 7 '15 at 16:55
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    $\begingroup$ Your question is very vague... I can formulate a hypothesis: perhaps ${\mathcal S}(G)$ is the subspace in $L_2(G)$ consisting of functions $f:G\to{\mathbb C}$ such that $f$ and its Fourier transform $\hat{f}$ rest in $L_2$ after being multiplied by any polynomial $p$ (where a polynomial on $G$ is a linear combination of finite poducts of real characters, i.e. continuous homomorphisms $r:G\to{\mathbb R}$). I don't know, whether this is more intuitively reasonable (and whether this is true). $\endgroup$ – Sergei Akbarov Sep 7 '15 at 18:51
  • $\begingroup$ Yes, this seems to be true, if we change a bit the definition of a polynomial: $p$ must be locally representable as a finite linear combination of finite products of real characters. $\endgroup$ – Sergei Akbarov Sep 7 '15 at 19:07
  • $\begingroup$ why use multiplication by characters instead of $|x|$ as Simon is suggesting? $\endgroup$ – Abdelmalek Abdesselam Sep 10 '15 at 14:37
  • $\begingroup$ For any LCA-group $G$ the multiplication by a real character on $\widehat{G}$ corresponds to the differentiation over a one-parametric subgroup in $G$, that's why I am speaking about polynomials. I don't know whether the multiplication by $|x|^k$ gives the same class of functions, I don't have intuition in $p$-adic fields. $\endgroup$ – Sergei Akbarov Sep 10 '15 at 20:36

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